Question

The sum of $(n-1)$ terms of $1+\left(1+3\right)+\left(1+3+5\right)+.........$ is

• A. $\frac{n\left(n+1\right)\left(2n+1\right)}{6}$
• B. $\frac{n^2\left(n+1\right)}{4}$
• C. $\frac{n\left(n-1\right)\left(2n-1\right)}{6}$
• D. $n^2$

Answer 1

Answer: (C)

$\frac{n\left(n-1\right)\left(2n-1\right)}{6}$

$S=1+(1+3)+(1+3+5)+...t_n=(1+3+5+....+2n-3)t_n=sumof(n-1)oddnumbers=(n-1{)}^{2}S=\sum t_n=\sum n^2-2\sum n+n\frac{n(n+1)(2n+1)}{6}-n(n+1)+n=\frac{n(n+1)(2n+1)}{6}-n^2=n(\frac{2n^2-3n+1}{6}-n)=\frac{n(2n^2-3n+1)}{6}S=\frac{n(n-1)(2n-1)}{6}$