Question

Find the sum of n terms of the series1² ⋅ 2 + 2²⋅3 + 3²⋅4 + 4²⋅5 +.....+up to n terms
1. \(\rm { \frac{n(n+1)(3n+1)(n+2)}{6}} \)
2. \(\rm { \frac{n(n+1)(3n+1)(n+2)}{12}} \)
3. \(\rm { \frac{n(n+1)(3n-1)(n+2)}{12}} \)
4. \(\rm { \frac{n(n+1)(3n+1)(n-2)}{12}} \)

Answer 1

Correct Answer - Option 2 : \(\rm { \frac{n(n+1)(3n+1)(n+2)}{12}} \)

Concept:

• Sum of the first n Natural Numbers \(\rm 1+2+3+4+....+n =\sum n = \frac{n(n+1)}{2}\)
• Sum of the Square of the first n Natural Numbers \(\rm 1^2+2^2+3^2+4^2+....+ n^2=\sum n^2 = \frac{n(n+1)(2n+1)}{6}\)
• Sum of the Cubes of the first n Natural Numbers \(\rm 1^3+2^3+3^3+4^3+....+ n^3=\sum n^3 = \frac{[n(n+1)]^2}{4}\)

 

Calculation:

Here, we have to find the sum of the series 12 ⋅ 2 + 22⋅3 + 32⋅4 + 42⋅5 +.....+up to n terms

nth term of the given series is T_n = n² ⋅ (n + 1) = n3 + n² 

Sum of n terms of the series,  \(\rm S_n = \sum T_n= \sum (n^3+n^2)=\sum n^3+\sum n^2\)

As we know that, 

\(\rm 1^2+2^2+3^2+4^2+....+ n^2=\sum n^2 = \frac{n(n+1)(2n+1)}{6}\) and \(\rm 1^3+2^3+3^3+4^3+....+ n^3=\sum n^3 = \frac{[n(n+1)]^2}{4}\)

\(\Rightarrow \rm S_n =\left ( { \frac{n(n+1)}{2}} \right )^2+\frac{n(n+1)(2n+1)}{6}\)

\(\Rightarrow \rm S_n =\left ( { \frac{n(n+1)}{2}} \right )\left ( \frac{n(n+1)}{2}+\frac{2n +1}{3} \right )\)

\(\Rightarrow \rm S_n =\left ( { \frac{n(n+1)}{2}} \right )\left ( \frac{3n^2+7n+2}{6} \right )\)

\(\Rightarrow \rm S_n =\left ( { \frac{n(n+1)}{2}} \right )\left ( \frac{(3n+1)(n+2)}{6} \right )\)

\(\Rightarrow \rm S_n =\left ( { \frac{n(n+1)(3n+1)(n+2)}{12}} \right )\)

Hence, option 2 is the correct answer.