The sum of the series 1 cdot 3 cdot 5 + 2 cdo | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The $n$-th term of the series is given by the product of three arithmetic progressions: $T_n = n(2n + 1)(3n + 2)$.Expanding this,we get $T_n = n(6

Vedclass · Accepted Answer

$\frac{n(n + 1)(9n^2 + 23n + 13)}{6}$

Vedclass · Answer

(A) The $n$-th term of the series is given by the product of three arithmetic progressions: $T_n = n(2n + 1)(3n + 2)$.Expanding this,we get $T_n = n(6n^2 + 4n + 3n + 2) = 6n^3 + 7n^2 + 2n$.The sum $S_n$ is $\sum_{k=1}^{n} T_k = 6\sum k^3 + 7\sum k^2 + 2\sum k$.Using standard summation formulas:$S_n = 6 \left[ \frac{n(n+1)}{2} \right]^2 + 7 \left[ \frac{n(n+1)(2n+1)}{6} \right] + 2 \left[ \frac{n(n+1)}{2} \right]$.Factoring out $\frac{n(n+1)}{6}$,we get:$S_n = \frac{n(n+1)}{6} [9n(n+1) + 7(2n+1) + 6]$.$S_n = \frac{n(n+1)}{6} [9n^2 + 9n + 14n + 7 + 6]$.$S_n = \frac{n(n+1)(9n^2 + 23n + 13)}{6}$.

The sum of the series $1 \cdot 3 \cdot 5 + 2 \cdot 5 \cdot 8 + 3 \cdot 7 \cdot 11 + \dots$ up to $n$ terms is:

Similar Questions

If $f(1)=3$,and $f(n+1)-f(n)=3(4^n-1)$,then for all $n \in N$,$f(n)=$

Let $S_{k} = \frac{1+2+\ldots+k}{k}$ and $\sum_{j=1}^n S_j^2 = \frac{n}{A}(Bn^2 + Cn + D)$,where $A, B, C, D \in \mathbb{N}$ and $A$ has the least value. Then:

If the $n^{th}$ term of a sequence is $T_n = 2n - 1$,then the sum of $n$ terms $S_n = \dots$

The value of $\sum\limits_{n = 0}^\infty {\frac{{{{(n + 1)}^2}}}{{{7^n}}}}$ is -

Find the $20^{\text{th}}$ term in the following sequence whose $n^{\text{th}}$ term is $a_{n} = \frac{n(n-2)}{n+3}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module