Question

To find:
$1^2+2^2+3^2...................+n^2$

• A. $=\frac{n^3}{2}$
• B. $>\frac{n^3}{2}$
• C. $>\frac{n^3}{3}$
• D. $=\frac{n^3}{3}$

Answer 1

The correct option is C $>\frac{n^3}{3}$

Let $n=1$
$\frac{n^3}{2}=\frac{1}{2};\frac{n^3}{3}=\frac{1}{3}$
Let $n=2,\Rightarrow 1^2+2^2=5$
$\frac{n^3}{2}=4;\frac{n^3}{3}=\frac{8}{3}$
Let$n=3,\Rightarrow 1^2+2^2+3^2=14$
$\frac{n^3}{2}=\frac{27}{2};\frac{n^3}{3}=9$
Let$n=4,\Rightarrow 1^2+2^2+3^2+4^2=30$
$\frac{n^3}{2}=32;\frac{n^3}{3}=\frac{64}{3}$
$1^2+2^2+3^2+\cdots +n^2>\frac{n^3}{3}$
$P\left(n\right):1^2+2^2+3^2+\cdots +n^2>\frac{n^3}{3}$
$P\left(1\right)$ is true
Let $P\left(k\right)$ be true.
$\Rightarrow 1^2+2^2+3^2+\cdots +k^2>\frac{k^3}{3}$
$1^2+2^2+3^2+\cdots +k^2+(k+1{)}^2>\frac{k^3}{3}+k^2+2k+1$
$=\frac{k^3+3k^2+6k+3}{3}$
$=\frac{k^3+3k^2+3k+1+3k+2}{3}$
$=\frac{(k+1{)}^3}{3}+k+\frac{2}{3}$
$1^2+2^2+3^2+\cdots +k^2+(k+1{)}^2>\frac{(k+1{)}^3}{3}$
$P(k+1)$ is true.
$P\left(n\right)$ is true $\forall n\in N$