InterviewSolution
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श्रेणी `1.2.3+2.3.4+3.4.5+...+n` पदों तक योगफल ज्ञात कीजिए | |
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Answer» यहाँ,`a_(n)=(1,2,3...का nवाँ पद)xx(2,3,4...का n वाँ पद)xx(3,4,5...का n वाँ पद)` `=(1+(n-1)xx1)xx1)xx(2+(n-1)xx1)xx(3+(n-1)xx1)` `=n(n+1)(n+2)=n^(3)+3n^(2)+2n` अब `" "S_(n)=underset(K=1)overset(n)Sigma a_(K)=underset(K=1)overset(n)Sigma (K^(3)+3K^(2)+2K)=underset(K=1)overset(n)Sigma K^(3)+3 underset(K=1)overset(n)Sigma K^(2)+2underset(K=1)overset(n)Sigma K` `=(1)/(4)n^(2)(n+1)^(2)+3.(1)/(6)+3.(1)/(6)n(n+1)(2n+1)+2.(1)/(2)n(n+1)` `=(1)/(4)n(n+1){n(n+1)+2(2n+1)+4}` `=(1)/(4)n(n+1)[(n^(2)+5n+6)]=(1)/(4)n(n+1)(n+2)(n+3)` जो कि अभीष्ट योगफल है | |
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