Question

$\frac{1^2}{1}+\frac{1^2+2^2}{1+2}+\frac{1^2+2^2+3^2}{1+2+3}+\dots +n$ terms $=$

• A. $\frac{n(n+3)}{4}$
• B. $\frac{n(n+3)}{5}$
• C. $\frac{n(n+2)}{3}$
• D. $\frac{n(n+5)}{6}$

Answer 1

Answer: (C)

$\frac{n(n+2)}{3}$

We Know,

${\sum }_{i=1}^{n}k=1+2+3+\dots +n=\frac{n(n+1)}{2}$ (sum of first n natural numbers)

${\sum }_{i=1}^n{k}^2=1^2+2^2+3^2+\dots +n^2=\frac{n(n+1)(2n+1)}{6}$ (sum of squares of the first n natural numbers)

$\therefore t_r=\frac{\Sigma k^2}{\Sigma k}=\frac{\frac{n(n+1)(2n+1)}{6}}{\frac{n(n+1)}{2}}=\frac{2n+1}{3}$

$S=\frac{2}{3}\Sigma n+\frac{1}{3}\Sigma 1$

$S=\frac{2}{3}\frac{n(n+1)}{2}+\frac{1}{3}n$

$S=\frac{n(n+2)}{3}$