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ગણિતીય અનુમાનનો સિદ્ધાંત

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ગણિત ધોરણ 11 સેમિસ્ટર 2

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ગણિત

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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. 


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