Question

The sum $1+\frac{1^3+2^3}{1+2}+\frac{1^3+2^3+3^3}{1+2+3}+\cdot \cdot \cdot +\frac{1^3+2^3+3^3+\cdot \cdot \cdot +{15}^3}{1+2+3+\cdot \cdot \cdot +15}-\frac{1}{2}(1+2+3+\cdot \cdot \cdot +15)$ is equal to :

• A. $620$
• B. $660$
• C. $1240$
• D. $1860$

Answer 1

The correct option is A $620$

Sum $=\sum _{n=1}^{15}\frac{1^3+2^3+\cdot \cdot \cdot +n^3}{1+2+\cdot \cdot \cdot +n}-\frac{1}{2}\frac{15\times 16}{2}$
$=\sum _{n=1}^{15}\frac{{\left(\frac{n(n+1)}{2}\right)}^2}{\frac{n(n+1)}{2}}-60$
$=\frac{1}{2}\sum _{n=1}^{15}{n}^2+\frac{1}{2}\sum _{n=1}^{15}n-60$
$=\frac{1}{2}\times \frac{15\times 16\times 31}{6}+\frac{1}{2}\times \frac{15\times 16}{2}-60$
$=620$