Latest UPSC Aptitude Question SOLUTION: Q. Find the sum of following series. 1*2+2*3+3*4+4*5+...upto n ?

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(#M40031778) UPSC Numerical Ability question Keep an EYE

Q. Find the sum of following series.
1*2+2*3+3*4+4*5+...upto n ?

Asked In UPSC Chandini Samal (10 years ago)

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RAKESH  (10 years ago)

1*2+2*3+3*4+4*5+...+n(n+1)
Tn=n(n+1)=n^2+n
Sum= Σ(Tn)=Σ(n^2+n)[for n=1 to n]
=Σ(n^2)+Σ(n)[for n=1 to n]
=n(n+1)*(2n+1)/6 + n(n+1)/2
=n(n+1)/2 *[(2n+1)/3 + 1]
=n(n+1)/2 *(2n+4)/3
=n(n+1)(n+2)/3
=1/3[n*(n+1)*(n+2)]

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DHAKAD NEETENDRA SINGH  (10 years ago)

nth term for this= n(n+1)
summation of n(n+1) terms is given by
= summation{n(n+1)}
= summation{n^2+n}
= summation n^2 + summation n
= {n(n+1)(2n+1)}/6 * n(n+1)/2

= n(n+1)/2 [ (2n+1)/3 + 1 ] taking common terms
= 1/3[n(n+1)(n+2)] ans.....

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