The sum to n terms of (2n - 1) + 2(2n - 3) + | vedclass

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

Question

(D) The $k$-th term of the series is $T_k = k(2n - (2k - 1)) = k(2n - 2k + 1) = 2nk - 2k^2 + k$.Summing from $k = 1$ to $n$:$S = \sum_{k=1}^{n} (2nk -

Vedclass · Accepted Answer

$n(n + 1)(2n + 1)/6$

Vedclass · Answer

(D) The $k$-th term of the series is $T_k = k(2n - (2k - 1)) = k(2n - 2k + 1) = 2nk - 2k^2 + k$.Summing from $k = 1$ to $n$:$S = \sum_{k=1}^{n} (2nk - 2k^2 + k) = 2n \sum_{k=1}^{n} k - 2 \sum_{k=1}^{n} k^2 + \sum_{k=1}^{n} k$.Using standard summation formulas:$S = 2n \frac{n(n+1)}{2} - 2 \frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2}$.$S = n^2(n+1) - \frac{n(n+1)(2n+1)}{3} + \frac{n(n+1)}{2}$.Factor out $\frac{n(n+1)}{6}$:$S = \frac{n(n+1)}{6} [6n - 2(2n+1) + 3] = \frac{n(n+1)}{6} [6n - 4n - 2 + 3] = \frac{n(n+1)(2n+1)}{6}$.

The sum to $n$ terms of $(2n - 1) + 2(2n - 3) + 3(2n - 5) + \dots$ is

Similar Questions

$\frac{1^{2}}{2} + \frac{1^{2}+2^{2}}{3} + \frac{1^{2}+2^{2}+3^{2}}{4} + \frac{1^{2}+2^{2}+3^{2}+4^{2}}{5} + \dots$ up to $8$ terms $=$

The sum of the series $1 + 3x + 6x^2 + 10x^3 + \dots \infty$ is

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

The sum $11^3 + 12^3 + \dots + 20^3$ is:

Find the sum of the series $1 + 3 + 7 + 15 + 31 + \dots$ up to $n$ terms.

Vedclass Test Series

Exam Paper Generator

Online Exam Module