Question

Sum of the series $1\left(1\right)+2(1+3)+3(1+3+5)+4(1+3+5+7)+....+10(1+3+5+7+....+19)$ is equal to

• A. $385$
• B. $1025$
• C. $1125$
• D. $2025$
• E. $3025$

Answer 1

The correct option is E $3025$

$1\left(1\right)+2(1+3)+3(1+3+5)+4(1+3+5+7)+....+10(1+3+5+7+....+19)$
$=1\times 1+2\times 2^2+3\times 3^2+4\times 4^2+....+\times {10}^2$
$={\left[\frac{10(10+1)}{2}\right]}^2$
[$\because$ sumof cubes of $n$ natural numbers $=\frac{n{(n+1)}^2}{2}$]
$={\left[\frac{10\times 11}{2}\right]}^2={\left(55\right)}^2=3025$