The sum of the series 1 + 2 times 3 + 3 times | vedclass

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

Question

(B) The $n^{th}$ term of the series is given by $T_n = n(2n - 1) = 2n^2 - n$.The sum of the first $n$ terms is $S_n = \sum_{k=1}^{n} (2k^2 - k) = 2 \s

Vedclass · Accepted Answer

$946$

Vedclass · Answer

(B) The $n^{th}$ term of the series is given by $T_n = n(2n - 1) = 2n^2 - n$.The sum of the first $n$ terms is $S_n = \sum_{k=1}^{n} (2k^2 - k) = 2 \sum_{k=1}^{n} k^2 - \sum_{k=1}^{n} k$.Using the standard summation formulas $\sum k^2 = \frac{n(n+1)(2n+1)}{6}$ and $\sum k = \frac{n(n+1)}{2}$,we get:$S_n = 2 \left[ \frac{n(n+1)(2n+1)}{6} ight] - \frac{n(n+1)}{2} = \frac{n(n+1)(2n+1)}{3} - \frac{n(n+1)}{2}$.For $n = 11$:$S_{11} = \frac{11(12)(23)}{3} - \frac{11(12)}{2}$.$S_{11} = 11 	imes 4 	imes 23 - 11 	imes 6$.$S_{11} = 1012 - 66 = 946$.

The sum of the series $1 + 2 \times 3 + 3 \times 5 + 4 \times 7 + \dots$ up to the $11^{th}$ term is:

Similar Questions

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

If $S_{1}, S_{2}, S_{3}$ are the sum of first $n$ natural numbers,their squares,and their cubes,respectively,show that $9 S_{2}^{2} = S_{3}(1 + 8 S_{1})$.

The expression for $a_n$ which satisfies $a_0=0, a_1=1$ and $a_n=a_{n-1}+a_{n-2}, \forall n \in N -\{0,1\}$ is:

Let $\langle a_n \rangle$ be a sequence such that $a_1+a_2+\ldots+a_n = \frac{n^2+3n}{(n+1)(n+2)}$. If $28 \sum_{k=1}^{10} \frac{1}{a_k} = p_1 p_2 p_3 \ldots p_m$,where $p_1, p_2, \ldots, p_m$ are the first $m$ prime numbers,then $m$ is equal to

Let $A$ be the sum of the first $20$ terms and $B$ be the sum of the first $40$ terms of the series $1^2 + 2 \cdot 2^2 + 3^2 + 2 \cdot 4^2 + 5^2 + \dots$. If $B - 2A = 100\lambda$,then $\lambda$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module