Question

If n is an odd integer greater than or equal to 1 then the value of $n^3-{(n-1)}^3+{(n-2)}^3-...+{(-1)}^{n-1}{.1}^3$ is

• A. $\frac{{(n+1)}^2.(2n-1)}{4}$
• B. $\frac{{(n-1)}^2.(2n-1)}{4}$
• C. $\frac{{(n+1)}^2.(2n+1)}{4}$
• D. none of these

Answer 1

The correct option is A $\frac{{(n+1)}^2.(2n-1)}{4}$

The sequence can be written as :
$S_n=n^3+{(n-2)}^3+{(n-4)}^3+......+1^3-[{(n-1)}^3+{(n-3)}^3+{(n-5)}^3+....+2^3]=n^3+{(n-2)}^3+{(n-4)}^3+......+1^3+[{(n-1)}^3+{(n-3)}^3+{(n-5)}^3+....+2^3]-2[{(n-1)}^3+{(n-3)}^3+{(n-5)}^3+....+2^3]={\sum }_{r=1}^n{r}^3-2\times 2^3[1^3+2^3+3^3+............+{\left(\frac{n-1}{2}\right)}^3]$
we have taken 8 common from the 2nd part of the sequence.
$={\sum }_{r=1}^n{r}^3-16{\sum }_1^{\frac{n-1}{2}}{r}^3=\left[{\frac{n(n+1)}{2}}^2\right]-16\left[{\frac{\frac{n-1}{2}(\frac{n-1}{2}+1)}{2}}^2\right]={\frac{n(n+1)}{2}}^2-16\left[{\frac{(n-1)(n+1)}{8}}^2\right]=\frac{{(n+1)}^2}{4}[n^2-{(n-1)}^2]=\frac{{(n+1)}^2(2n-1)}{4}$