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

ગણિતીય અનુમાનનો સિદ્ધાંત

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

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.