The value of the sum 1 times 2 times 3 + 2 ti | vedclass

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

Question

(D) The $n$-th term of the series is $T_n = n(n+1)(n+2)$.Expanding this,we get $T_n = n(n^2 + 3n + 2) = n^3 + 3n^2 + 2n$.The sum of $n$ terms is $S_n

Vedclass · Accepted Answer

$\frac{1}{4} n(n+1)(n+2)(n+3)$

Vedclass · Answer

(D) The $n$-th term of the series is $T_n = n(n+1)(n+2)$.Expanding this,we get $T_n = n(n^2 + 3n + 2) = n^3 + 3n^2 + 2n$.The sum of $n$ terms is $S_n = \sum_{k=1}^n T_k = \sum_{k=1}^n (k^3 + 3k^2 + 2k)$.Using standard summation formulas:$\sum_{k=1}^n k^3 = \frac{n^2(n+1)^2}{4}$,$\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$,and $\sum_{k=1}^n k = \frac{n(n+1)}{2}$.Substituting these:$S_n = \frac{n^2(n+1)^2}{4} + 3 \cdot \frac{n(n+1)(2n+1)}{6} + 2 \cdot \frac{n(n+1)}{2}$$S_n = \frac{n(n+1)}{2} \left[ \frac{n(n+1)}{2} + (2n+1) + 2 \right]$$S_n = \frac{n(n+1)}{2} \left[ \frac{n^2+n+4n+2+4}{2} \right] = \frac{n(n+1)(n^2+5n+6)}{4}$Since $n^2+5n+6 = (n+2)(n+3)$,we have $S_n = \frac{n(n+1)(n+2)(n+3)}{4}$.

The value of the sum $1 \times 2 \times 3 + 2 \times 3 \times 4 + 3 \times 4 \times 5 + \ldots$ up to $n$ terms is equal to

Similar Questions

The sum of the first $9$ terms of the series $\frac{1^3}{1} + \frac{1^3 + 2^3}{1 + 3} + \frac{1^3 + 2^3 + 3^3}{1 + 3 + 5} + \dots$ is:

For a series whose $n^{th}$ term is $\left( \frac{n}{x} \right) + y$,the sum of $r$ terms is:

If $2.5+5.9+8.13+11.17+\ldots$ to $n$ terms $=an^3+bn^2+cn+d$,then $a-b+c-d=$

For a series $S = 1 - 2 + 3 - 4 + \dots$ up to $n$ terms,
Statement-$1$: The sum of the series is always dependent on the value of $n$,i.e.,whether it is even or odd.
Statement-$2$: The sum of the series is $-\frac{n}{2}$ when the value of $n$ is any even integer.

$\sum\limits_{n = 1}^\infty {\sum\limits_{k = 1}^{n - 1} {\frac{k}{{{2^{n + k}}}}} } $ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module