Find the sum of n terms of the series 1 + 6 + | vedclass

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

Question

(B) Let the series be $S_n = 1 + 6 + 13 + 22 + 33 + \dots + T_n$. The differences between consecutive terms are $5, 7, 9, 11, \dots$,which form an Ari

Vedclass · Accepted Answer

$\frac{n}{6}[2n^2 + 9n - 5]$

Vedclass · Answer

(B) Let the series be $S_n = 1 + 6 + 13 + 22 + 33 + \dots + T_n$. The differences between consecutive terms are $5, 7, 9, 11, \dots$,which form an Arithmetic Progression $(AP)$. Let $T_n = an^2 + bn + c$. For $n=1, T_1 = a + b + c = 1$. For $n=2, T_2 = 4a + 2b + c = 6$. For $n=3, T_3 = 9a + 3b + c = 13$. Subtracting the equations: $(4a + 2b + c) - (a + b + c) = 6 - 1 \implies 3a + b = 5$. $(9a + 3b + c) - (4a + 2b + c) = 13 - 6 \implies 5a + b = 7$. Subtracting these results: $(5a + b) - (3a + b) = 7 - 5 \implies 2a = 2 \implies a = 1$. Substituting $a=1$ into $3a + b = 5$,we get $3(1) + b = 5 \implies b = 2$. Substituting $a=1, b=2$ into $a + b + c = 1$,we get $1 + 2 + c = 1 \implies c = -2$. Thus,$T_n = n^2 + 2n - 2$. Now,$S_n = \sum_{k=1}^n T_k = \sum_{k=1}^n (k^2 + 2k - 2) = \sum k^2 + 2\sum k - \sum 2$. $S_n = \frac{n(n+1)(2n+1)}{6} + 2 \cdot \frac{n(n+1)}{2} - 2n$. $S_n = \frac{n(2n^2 + 3n + 1) + 6n(n+1) - 12n}{6} = \frac{n(2n^2 + 3n + 1 + 6n + 6 - 12)}{6} = \frac{n(2n^2 + 9n - 5)}{6}$.

Find the sum of $n$ terms of the series $1 + 6 + 13 + 22 + 33 + \dots$

Similar Questions

The number of positive integers $n$ in the set $\{1, 2, 3, \ldots, 100\}$ for which the number $\frac{1^2+2^2+3^2+\ldots+n^2}{1+2+3+\ldots+n}$ is an integer is

The sum to infinity of the progression $9 - 3 + 1 - \frac{1}{3} + \dots$ is

The sum of infinite terms of the geometric progression $\frac{\sqrt{2} + 1}{\sqrt{2} - 1}, \frac{1}{2 - \sqrt{2}}, \frac{1}{2}, \dots$ is

Let $a_n$ be a sequence such that $a_1 = 5$ and $a_{n+1} = a_n + (n - 2)$ for all $n \in N$. Then $a_{51}$ is:

The sum of $n$ terms of the series whose $n^{th}$ term is $n(n + 1)$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module