Find the sum of the series 2 + 4 + 7 + 11 + 1 | vedclass

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

Question

(B) Let the series be $S_n = 2 + 4 + 7 + 11 + 16 + \dots + T_n$. The differences between consecutive terms are $2, 3, 4, 5, \dots$,which form an arith

Vedclass · Accepted Answer

$\frac{n}{6}(n^2 + 3n + 8)$

Vedclass · Answer

(B) Let the series be $S_n = 2 + 4 + 7 + 11 + 16 + \dots + T_n$. The differences between consecutive terms are $2, 3, 4, 5, \dots$,which form an arithmetic progression. Let $T_n = an^2 + bn + c$. For $n=1, T_1 = a + b + c = 2$. For $n=2, T_2 = 4a + 2b + c = 4$. For $n=3, T_3 = 9a + 3b + c = 7$. Subtracting the equations,we get $3a + b = 2$ and $5a + b = 3$. Solving these,$2a = 1 \implies a = 1/2$,$b = 1/2$,and $c = 1$. Thus,$T_n = \frac{1}{2}n^2 + \frac{1}{2}n + 1 = \frac{n^2 + n + 2}{2}$. The sum $S_n = \sum_{k=1}^n T_k = \frac{1}{2} [\sum k^2 + \sum k + \sum 2]$. $S_n = \frac{1}{2} [\frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2} + 2n]$. $S_n = \frac{n}{12} [(n+1)(2n+1) + 3(n+1) + 12] = \frac{n}{12} [2n^2 + 3n + 1 + 3n + 3 + 12] = \frac{n}{12} [2n^2 + 6n + 16] = \frac{n}{6} (n^2 + 3n + 8)$.

Find the sum of the series $2 + 4 + 7 + 11 + 16 + \dots$ up to $n$ terms.

Similar Questions

If $1+\frac{\cos \theta}{2}+\frac{\cos 2 \theta}{4}+\frac{\cos 3 \theta}{8}+\ldots = \frac{a-2 \cos \theta}{5+b \cos \theta}$ for some $a, b \in R$,then $(a-b)^2=$

In the sequence of sets $(1,2,3), (4,5,6), (7,8,9,10), \ldots$,the sum of elements in the $50^{th}$ set is

What is the sum of the first $20$ terms of the sequence $0.7, 0.77, 0.777, \dots$?

Show that $\frac{1 \times 2^{2}+2 \times 3^{2}+\ldots+n \times(n+1)^{2}}{1^{2} \times 2+2^{2} \times 3+\ldots+n^{2} \times(n+1)}=\frac{3 n+5}{3 n+1}$

The sum of all positive integers $n$ for which $\frac{1^3+2^3+\ldots+(2n)^3}{1^2+2^2+\ldots+n^2}$ is also an integer is

Vedclass Test Series

Exam Paper Generator

Online Exam Module