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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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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.







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