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. $\frac{(n+1{)}^2(2n+1)}{8}$

Answer 1

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

The given series is

$S=1^3-2^3+3^3-4^3+5^3-6^3+...-(n-1{)}^3+n^3$

$\because n$ is an odd integer.

This is the same as

$S=(1^3+2^3+3^3+...+n^3)-2(2^3+4^3+6^3+...+(n-1{)}^3)$

Which can be written as

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

Which can now be solved using

$\sum r^3=\frac{n^2(n+1{)}^2}{4}$

$\therefore$ The desired answer is

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

Hence, option A is the correct answer.