Question

$\frac{1^3+2^3+3^3+....+{12}^3}{1^2+2^2+3^2+....+{12}^2}=$

• A. $\frac{234}{25}$
• B. $\frac{234}{35}$
• C. $\frac{263}{27}$
• D. $noneofthese$

Answer 1

The correct option is A

$\frac{234}{25}$

Explanation for the correct option:

As we know,

Sum of cubes of $n$ natural numbers $={\left(\frac{n(n+1)}{2}\right)}^2$

Sum of squares $n$ natural numbers $=\frac{n(n+1)(2n+1)}{6}$

Thus $\frac{1^3+2^3+3^3+....+n^3}{1^2+2^2+3^2+....+n^2}=\frac{{\left({\displaystyle \frac{n(n+1)}{2}}\right)}^2}{\left[{\displaystyle \frac{n(n+1)(2n+1)}{6}}\right]}$

$=\frac{3n(n+1)}{2(2n+1)}$

Put $n=12$ in above equation:

$\frac{1^3+2^3+3^3+....+{12}^3}{1^2+2^2+3^2+....+{12}^2}=\frac{3\times 12\times 13}{2\times 25}$

$=\frac{\mathbf{234}}{\mathbf{25}}$

Hence, Option ‘A’ is Correct.