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 $\frac{(n+1{)}^2(2n-1)}{4}$

As $n$ odd, the value of the given expression

$=1^3-2^3+3^3-...+n^3$

$=(1^3+2^3+3^3+...+n^3)-2(2^3+4^3+...{(n-1)}^3)$

$={\left\{\frac{n(n+1)}{2}\right\}}^2-16\{1^3+2^3+...+{\left(\frac{n-1}{2}\right)}^3\}$

$=\frac{n^2{(n+1)}^2}{4}-16.{\left\{\frac{\frac{n-1}{2}.\frac{n+1}{2}}{2}\right\}}^2.$

$=\frac{{(n+1)}^2(2n-1)}{4}$