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Question

Mathematics Question on Arithmetic Progression

The sum of 1st1^{st} nn terms of the series 121+12+221+2+12+22+321+2+3+...\frac{1^{2}}{1}+\frac{1^{2}+2^{2}}{1+2}+\frac{1^{2}+2^{2}+3^{2}}{1+2+3} + ...

A

\cen+23\ce{\frac{n + 2}{3}}

B

\cen(n+2)3\ce{\frac{n(n + 2)}{3}}

C

\cen(n2)3\ce{\frac{n(n - 2)}{3}}

D

\cen(n2)6\ce{\frac{n(n - 2)}{6}}

Answer

\cen(n+2)3\ce{\frac{n(n + 2)}{3}}

Explanation

Solution

We have, 121+12+221+2+12+22+321+2+3+\frac{1^{2}}{1}+\frac{1^{2}+2^{2}}{1+2}+\frac{1^{2}+2^{2}+3^{2}}{1+2+3}+\ldots an=12+22+32++n21+2+3++n\therefore a_{n}=\frac{1^{2}+2^{2}+3^{2}+\ldots+n^{2}}{1+2+3+\ldots+n} an=n(n+1)(2n+1)6n(n+1)2\Rightarrow a_{n}=\frac{\frac{n(n+1)(2 n+1)}{6}}{\frac{n(n+1)}{2}} [Σn2=n(n+1)(2n+1)6andΣn=n(n+1)2=2n+13]\Rightarrow\left[\because \Sigma n^{2}=\frac{n(n+1)(2 n+1)}{6} \text{and} \, \Sigma n=\frac{n(n+1)}{2} \therefore =\frac{2 n+1}{3}\right] an=2n+13a_n = \frac{2n + 1}{3} Now, Sn=Σan=Σ2n+13=13[2n+Σ1]S_{n} =\Sigma a_{n}=\Sigma \frac{2 n+1}{3}=\frac{1}{3}\left[\sum 2 n+\Sigma 1\right] =13[2n(n+1)2+n][Σ1=n]=\frac{1}{3}\left[\frac{2 n(n+1)}{2}+n\right] \,\,[\because \Sigma 1=n] =13[n2+n+n]=\frac{1}{3}\left[n^{2}+n+n\right] =13(n2+2n)=n(n+2)3=\frac{1}{3}\left(n^{2}+2 n\right)=\frac{n(n+2)}{3}