Sequences & Series
Sum of Series
MMTS_Full_Test_04
Grade 12
Question:
Sum of the series $1^2\cdot2^2+2^2\cdot3^2+\cdots+n^2\cdot(n+1)^2$ is
$\dfrac{n(n+1)(n+2)(3n+1)}{12}$
$\dfrac{n(n+1)^2(n+2)}{4}$
$\dfrac{n(n+1)(n+2)(3n+5)}{12}$
$\dfrac{n(n+1)(2n+1)(3n^2+3n-1)}{30}$
Step-by-Step Solution
Key Concept: $r^2(r+1)^2=r^4+2r^3+r^2$; sum each
$\sum r^2(r+1)^2=\sum(r^4+2r^3+r^2)=\frac{n(n+1)(2n+1)(3n^2+3n-1)}{30}+\frac{n^2(n+1)^2}{2}+\frac{n(n+1)(2n+1)}{6}$. Simplifies to $\frac{n(n+1)(2n+1)(3n^2+3n-1)}{30}$.
Correct Answer: 3