Chapter Chosen

ગણિતીય અનુમાનનો સિદ્ધાંત

Book Chosen

ગણિત ધોરણ 11 સેમિસ્ટર 2

Subject Chosen

ગણિત

Book Store

Download books and chapters from book store.
Currently only available for
CBSE Gujarat Board Haryana Board

Previous Year Papers

Download the PDF Question Papers Free for off line practice and view the Solutions online.
Currently only available for
Class 10 Class 12



માટે P (1) સત્ય છે.

ધારો કે, P(k) સત્ય છે.

therefore space fraction numerator 1 over denominator 1 times 4 end fraction plus fraction numerator 1 over denominator 4 times 7 end fraction plus fraction numerator 1 over denominator 7 times 10 end fraction plus.... plus fraction numerator 1 over denominator open parentheses 3 straight k minus 2 close parentheses open parentheses 3 straight k plus 1 close parentheses end fraction equals space fraction numerator straight k over denominator 3 straight k plus 1 end fraction comma space straight k space element of space straight N space....... left parenthesis 1 right parenthesis


હવે, P ( k + 1 ) સત્ય બતાવવા માટે n = k + 1 લેતાં,

straight L. straight H. straight S. space equals space fraction numerator 1 over denominator 1 times 4 end fraction plus fraction numerator 1 over denominator 4 times 7 end fraction plus fraction numerator 1 over denominator 7 times 10 end fraction plus.... plus fraction numerator 1 over denominator open parentheses 3 straight k minus 2 close parentheses open parentheses 3 straight k plus 1 close parentheses end fraction plus fraction numerator 1 over denominator open parentheses 3 open parentheses straight k plus 1 close parentheses minus 2 close parentheses open parentheses 3 open parentheses straight k plus 1 close parentheses plus 1 close parentheses end fraction

space
space space space space space space space space space space space space space equals space fraction numerator straight k over denominator 3 straight k plus 1 end fraction plus fraction numerator 1 over denominator open parentheses 3 open parentheses straight k plus 1 close parentheses minus 2 close parentheses open parentheses 3 open parentheses straight k plus 1 close parentheses plus 1 close parentheses end fraction space space space space open square brackets because space left parenthesis 1 right parenthesis space પરથ ી space close square brackets


space space space space space space space space space space space space space equals space fraction numerator straight k over denominator 3 straight k plus 1 end fraction plus fraction numerator 1 over denominator open parentheses 3 straight k plus 1 close parentheses open parentheses 3 straight k plus 4 close parentheses end fraction


space space space space space space space space space space space space equals space fraction numerator straight k open parentheses 3 straight k plus 4 close parentheses plus 1 over denominator open parentheses 3 straight k plus 1 close parentheses open parentheses 3 straight k plus 4 close parentheses end fraction space


         equals space fraction numerator 3 straight k squared plus 4 straight k plus 1 over denominator open parentheses 3 straight k plus 1 close parentheses open parentheses 3 straight k plus 4 close parentheses end fraction


equals space fraction numerator open parentheses 3 straight k plus 1 close parentheses open parentheses straight k plus 1 close parentheses over denominator open parentheses 3 straight k plus 1 close parentheses open parentheses 3 straight k plus 4 close parentheses end fraction


equals space fraction numerator straight k plus 1 over denominator 3 straight k plus 4 end fraction


equals space fraction numerator straight k plus 1 over denominator 3 open parentheses straight k plus 1 close parentheses plus 1 end fraction space equals space straight R. straight H. straight S.



therefore space straight P space open parentheses straight k plus 1 close parentheses therefore space straight P open parentheses straight k close parentheses space સત ્ ય space છ ે. space rightwards double arrow space straight P space open parentheses straight k plus 1 close parentheses space સત ્ ય space છ ે.

વળી, P(1) પણ સત્ય છે.

તેથી ગણિતીય અનુમાનના સિદ્ધાંતથી P(n), દરેક n space element of space N માટે સત્ય છે.







1 times 2 times 3 space plus space 2 times 3 times 4 plus.... plus straight n open parentheses straight n plus 1 close parentheses open parentheses straight n plus 2 close parentheses space equals space fraction numerator straight n open parentheses straight n plus 1 close parentheses open parentheses straight n plus 2 close parentheses open parentheses straight n plus 3 close parentheses over denominator 4 end fraction

અહીં, straight P open parentheses straight n close parentheses space colon space 1 times 2 times 3 space plus space 2 times 3 times 4 plus 3 times 4 times 5 plus.... plus straight n open parentheses straight n plus 1 close parentheses open parentheses straight n plus 2 close parentheses space equals space fraction numerator straight n open parentheses straight n plus 1 close parentheses open parentheses straight n plus 2 close parentheses open parentheses straight n plus 3 close parentheses over denominator 4 end fraction space comma space straight n space element of space straight N

હવે, n = 1 માટે,

L. H. S. space equals space 1 times 2 times 3 space equals space 6 space space અન ે space R. H. S. space equals space fraction numerator 1 open parentheses 2 close parentheses open parentheses 3 close parentheses open parentheses 4 close parentheses over denominator 4 end fraction space equals space 6

therefore space L. H. S. space equals space R. H. S.


therefore space straight P open parentheses 1 close parentheses space સત્ય છે.

ધારો કે, P(k) સત્ય છે.

therefore space 1 times 2 times 3 plus 2 times 3 times 4 plus 3 times 4 times 5 plus.... plus k open parentheses k plus 1 close parentheses open parentheses k plus 2 close parentheses space equals space fraction numerator k open parentheses k plus 1 close parentheses open parentheses k plus 2 close parentheses open parentheses k plus 3 close parentheses over denominator 4 end fraction comma space n space element of space N space..... left parenthesis 1 right parenthesis

હવે, P ( k + 1 ) સત્ય બતાવવા માટે n = k + 1 લેતાં,


L. H. S. space equals space 1 times 2 times 3 space plus space 2 times 3 times 4 plus 3 times 4 times 5 plus.... plus k open parentheses k plus 1 close parentheses open parentheses k plus 2 close parentheses plus open parentheses k plus 1 close parentheses open parentheses k plus 2 close parentheses open parentheses k plus 3 close parentheses

space space space space space space space space space space space space space space equals space fraction numerator k open parentheses k plus 1 close parentheses open parentheses k plus 2 close parentheses open parentheses k plus 3 close parentheses over denominator 4 end fraction plus open parentheses k plus 1 close parentheses open parentheses k plus 2 close parentheses open parentheses k plus 3 close parentheses space space space open square brackets because space left parenthesis 1 right parenthesis space space પરથ ી close square brackets


          equals space open parentheses straight k plus 1 close parentheses open parentheses straight k plus 2 close parentheses open parentheses straight k plus 3 close parentheses open parentheses straight k over 4 plus 1 close parentheses

equals space fraction numerator open parentheses straight k plus 1 close parentheses open parentheses straight k plus 2 close parentheses open parentheses straight k plus 3 close parentheses open parentheses straight k plus 4 close parentheses over denominator 4 end fraction

equals space fraction numerator open parentheses straight k plus 1 close parentheses open parentheses open parentheses straight k plus 1 close parentheses plus 1 close parentheses open parentheses open parentheses straight k plus 1 close parentheses plus 2 close parentheses open parentheses open parentheses straight k plus 1 close parentheses plus 3 close parentheses over denominator 4 end fraction

equals space straight R. straight H. straight S.



તેથી P ( k + 1 ) સત્ય છે.

therefore space straight P open parentheses straight k close parentheses space સત ્ ય space છ ે. space rightwards double arrow space straight P space open parentheses straight k plus 1 close parentheses space સત ્ ય space છ ે.

વળી, P (1) પણ સત્ય છે.

માટે ગણિતીય સિદ્ધાંતથી P (n), દરેક n space element of space N space માટે સત્ય છે.



straight a plus open parentheses straight a plus straight d close parentheses plus open parentheses straight a plus 2 straight d close parentheses plus.... plus open parentheses straight a plus open parentheses straight n minus straight a close parentheses straight d close parentheses space equals space 1 half straight n open parentheses 2 straight a plus open parentheses straight n minus 1 close parentheses straight d close parentheses

અહ ીં space comma space straight P open parentheses straight n close parentheses space colon space straight a plus open parentheses straight a plus straight d close parentheses plus open parentheses straight a plus 2 straight d close parentheses plus open parentheses straight a plus 3 straight d close parentheses plus.... plus open square brackets straight a plus open parentheses straight n minus straight a close parentheses straight d close square brackets space equals space straight n over 2 open square brackets 2 straight a plus open parentheses straight n minus 1 close parentheses straight d close square brackets comma space straight n space element of space straight N

હવે, n = 1 માટે

L.H.S. = a અને R.H.S. = 1 half open square brackets 8 a plus open parentheses 1 minus 1 close parentheses d close square brackets space equals space a

therefore space straight L. straight H. straight S. space equals space straight R. straight H. straight S.

માટે, P (1) સત્ય છે.

therefore space straight a plus open parentheses straight a plus straight d close parentheses plus open parentheses straight a plus 2 straight d close parentheses plus open parentheses straight a plus 3 straight d close parentheses plus.... plus space open square brackets straight a plus open parentheses straight k minus 1 close parentheses straight d close square brackets space equals space straight k over 2 open square brackets 2 straight a plus open parentheses straight k minus 1 close parentheses straight d close square brackets comma space straight k space element of space straight N space....... left parenthesis 1 right parenthesis

હવે, P (k + 1) સત્ય બતાવવા માટે n = k + 1 લેતાં,

L. H. S. space equals space a plus open parentheses a plus d close parentheses plus open parentheses a plus 2 d close parentheses plus open parentheses a plus 3 d close parentheses plus space... space plus open square brackets a plus open parentheses k minus 1 close parentheses d close square brackets plus open parentheses a minus k d close parentheses

space space space space space space space space space space space space space space equals space k over 2 open square brackets 2 a plus open parentheses k minus 1 close parentheses d close square brackets plus open parentheses a plus k d close parentheses space space space space space open square brackets because space left parenthesis 1 right parenthesis space space પરથ ી space close square brackets

space space space space space space space space space space space space space space equals space k a plus k over 2 open parentheses k minus 1 close parentheses d plus a plus k d

space space space space space space space space space space space space space space equals space k a plus a plus k over 2 open parentheses k minus 1 close parentheses d plus k d

 equals space open parentheses straight k plus 1 close parentheses straight a plus fraction numerator straight k squared straight d over denominator 2 end fraction minus kd over 2 plus kd

equals open parentheses straight k plus 1 close parentheses straight a plus fraction numerator straight k squared straight d over denominator 2 end fraction plus kd over 2

equals space fraction numerator straight k plus 1 over denominator 2 end fraction times 2 straight a plus fraction numerator straight k plus 1 over denominator 2 end fraction times kd

equals space fraction numerator straight k plus 1 over denominator 2 end fraction open square brackets 2 a plus k d close square brackets

equals space fraction numerator k plus 1 over denominator 2 end fraction open square brackets 2 a plus open parentheses open parentheses k plus 1 close parentheses minus 1 close parentheses d close square brackets space equals space R. H. S.


therefore space straight P space open parentheses straight k plus 1 close parentheses સત્ય છે.

therefore space straight P open parentheses straight k close parentheses સત્ય છે. rightwards double arrow space straight P open parentheses straight k plus 1 close parentheses સત્ય છે.

વળી, P (1) પણ સત્ય છે.

તેથી ગણિતીય અનુમાનના સિદ્ધાંતથી P(n), દરેક n space element of space N માટે સત્ય છે.

         



1 plus 5 plus 9 plus.... plus open parentheses 4 n minus 3 close parentheses space equals space n open parentheses 2 n minus 1 close parentheses

અહીં, P (n) : 1 plus 5 plus 9 plus.... plus open parentheses 4 n minus 3 close parentheses space equals space n open parentheses 2 n minus 1 close parentheses comma space n space element of space N

હવે, n = 1 માટે,

straight L. straight H. straight S. space equals space 1 space અન ે space straight R. straight H. straight S. space equals space 1 space open parentheses 2 space times 1 minus 1 close parentheses space equals space 1

therefore space straight L. straight H. straight S. space equals space straight R. straight H. straight S.


therefore space straight P open parentheses 1 close parentheses સત્ય છે. 

therefore space 1 plus 5 plus 9 plus.... plus open parentheses 4 straight k minus 3 close parentheses space equals space straight k open parentheses 2 straight k minus 1 close parentheses space comma space straight k space element of space straight N space........ space left parenthesis 1 right parenthesis

હવે, P ( k + 1) સત્ય વતાવવા માટે n = k + 1 લેતાં,

straight L. straight H. straight S. space equals space 1 plus 5 plus 9 plus.... plus open parentheses 4 straight k minus 3 close parentheses plus open parentheses 4 open parentheses straight k plus 1 close parentheses minus 3 close parentheses

space space space space space space space space space space space space space equals space straight k open parentheses 2 straight k minus 1 close parentheses plus open parentheses 4 straight k plus 1 close parentheses space space open square brackets because space open parentheses 1 close parentheses space પરથ ી close square brackets

space space space space space space space space space space space space space equals space 2 straight k squared plus 3 straight k plus 1

space space space space space space space space space space space space space equals space open parentheses straight k plus 1 close parentheses open parentheses 2 straight k plus 1 close parentheses equals open parentheses straight k plus 1 close parentheses open parentheses 2 open parentheses straight k plus 1 close parentheses minus 1 close parentheses equals space straight R. straight H. straight S.




P ( k + 1 ) સત્ય છે.

P ( k ) સત્ય છે.rightwards double arrow P ( k + 1 ) સત્ય છે.

વળી, P (1) પણ સત્ય છે.

ગણિતીય અનુમાનના સિદ્ધાંતથી P(n), દરેક straight n space element of space straight N માટે સત્ય છે.


1 squared space times space 2 space plus space 2 squared space times space 3 space plus space... space plus space straight n squared space open parentheses straight n plus 1 close parentheses space equals space fraction numerator straight n open parentheses straight n plus 1 close parentheses open parentheses straight n plus 2 close parentheses open parentheses 3 straight n plus 1 close parentheses over denominator 12 end fraction

અહીં, straight P open parentheses straight n close parentheses space colon 1 squared space times 2 space plus space 2 squared times 3 space plus space.... space plus space straight n squared open parentheses straight n plus 1 close parentheses space equals space fraction numerator straight n open parentheses straight n plus 1 close parentheses open parentheses straight n plus 2 close parentheses open parentheses 3 straight n plus 1 close parentheses over denominator 12 end fraction comma space straight n space element of space straight N


હવે, n = 1 માટે,

L. H. S space equals space 1 squared space times space 2 space equals space 2 space space અન ે space R. H. S space equals space fraction numerator 1 open parentheses 1 plus 1 close parentheses open parentheses 1 plus 2 close parentheses open parentheses 3 times 1 plus 1 close parentheses over denominator 12 end fraction space equals space 2

therefore space L. H. S space equals space R. H. S 


therefore space straight P open parentheses 1 close parentheses space સત ્ ય space છ ે.

ધારો કે, P (k) સત્ય છે.

therefore space 1 squared space times space 2 space plus 2 squared times 3 space plus space.... space plus space straight k squared space open parentheses straight k plus 1 close parentheses space equals space fraction numerator straight k open parentheses straight k plus 1 close parentheses open parentheses straight k plus 2 close parentheses open parentheses 3 straight k plus 1 close parentheses over denominator 12 end fraction comma space straight k element of straight N space........ left parenthesis 1 right parenthesis

હવે, P (k + 1) સત્ય બતાવવા માટે n = k + 1 લેતાં, 


L. H. S space equals space 1 squared space times space 2 space plus space 2 squared space times 3 space plus space... space plus space k squared space open parentheses k plus 1 close parentheses plus open parentheses k plus 1 close parentheses squared open parentheses open parentheses k plus 1 close parentheses plus 1 close parentheses


space space space space space space space space space space space space equals space fraction numerator k open parentheses k plus 1 close parentheses open parentheses k plus 2 close parentheses open parentheses 3 k plus 1 close parentheses over denominator 12 end fraction plus open parentheses k plus 1 close parentheses squared open parentheses k plus 2 close parentheses space space space space space open square brackets because space left parenthesis 1 right parenthesis space પરથ ી close square brackets


space space space space space space space space space space space space equals space open parentheses K plus 1 close parentheses open parentheses K plus 2 close parentheses open square brackets fraction numerator K open parentheses 3 K plus 1 close parentheses plus 12 open parentheses K plus 1 close parentheses over denominator 12 end fraction close square brackets



equals space fraction numerator open parentheses straight K plus 1 close parentheses open parentheses straight K plus 2 close parentheses open parentheses 3 straight K squared plus 13 straight K plus 12 close parentheses over denominator 12 end fraction

equals space fraction numerator open parentheses straight K plus 1 close parentheses open parentheses straight K plus 2 close parentheses open parentheses straight K plus 3 close parentheses open parentheses 3 straight K plus 4 close parentheses over denominator 12 end fraction

equals space fraction numerator open parentheses straight K plus 1 close parentheses open parentheses open parentheses straight K plus 1 close parentheses plus 1 close parentheses open parentheses open parentheses straight K plus 1 close parentheses plus 2 close parentheses open parentheses 3 open parentheses straight K plus 1 close parentheses plus 1 close parentheses over denominator 12 end fraction

equals space straight R. straight H. straight S.


માટે P ( k + 1 ) સત્ય છે.

માટે P (k) સત્ય છે. P ( k + 1 ) સત્ય છે.

વળી, P(1) પણ સત્ય છે.

તો ગણિતીય અનુમાનના સિદ્ધાંતથી P (n), દરેક n space element of space N  માટે સત્ય છે.