Question

What is the sum of $1^3+2^3+3^3+....+n^3$ ?

• A. ${\left[\frac{n\left(n+1\right)}{3}\right]}^2$
• B. ${\left[\frac{n\left(n+1\right)}{2}\right]}^2$
• C. ${\left[\frac{n\left(n+2\right)}{3}\right]}^2$
• D. ${\left[\frac{n\left(n+1\right)}{2}\right]}^3$

Answer 1

Answer: (B)

$P\left(k\right)=1^3+2^3+3^3+...+k^3={\left\{\frac{k\left(k+1\right)}{2}\right\}}^2$
Now, $\left(1^3+2^3+3^3+...+k^3\right)+{\left(k+1\right)}^3$
$={\left\{\frac{k\left(k+1\right)}{2}\right\}}^2+{\left(k+1\right)}^3$
$=\frac{1}{4}{k}^2{\left(k+1\right)}^2+{\left(k+1\right)}^3$
$=\frac{1}{4}\left[k^2{\left(k+1\right)}^2+4{\left(k+1\right)}^3\right]$
$=\frac{1}{4}{\left(k+1\right)}^2\left[k^2+4\left(k+1\right)\right]$
$=\frac{1}{4}{\left(k+1\right)}^2+{\left(k+2\right)}^2$