If S_n = 2 + 4 + 7 + 11 + dots + n terms,then | vedclass

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

Question

(C) The given series is $S_n = 2 + 4 + 7 + 11 + \dots + t_n$.Let the sequence be $a_n = 2, 4, 7, 11, \dots$.The differences between consecutive terms

Vedclass · Accepted Answer

$\frac{n^2 + n + 2}{2}$

Vedclass · Answer

(C) The given series is $S_n = 2 + 4 + 7 + 11 + \dots + t_n$.Let the sequence be $a_n = 2, 4, 7, 11, \dots$.The differences between consecutive terms are $2, 3, 4, \dots$,which form an arithmetic progression.Thus,$t_n = t_1 + \sum_{k=1}^{n-1} (k+1)$.$t_n = 2 + [2 + 3 + 4 + \dots + n]$.This is $t_n = 1 + [1 + 2 + 3 + \dots + n]$.Using the sum formula $\sum_{k=1}^{n} k = \frac{n(n+1)}{2}$,we get:$t_n = 1 + \frac{n(n+1)}{2} = \frac{2 + n^2 + n}{2} = \frac{n^2 + n + 2}{2}$.

If $S_n = 2 + 4 + 7 + 11 + \dots + n$ terms,then $t_n = \dots$

Similar Questions

If $\sum\limits_{i = 1}^n i = \frac{n(n + 1)}{2}$,then $\sum\limits_{i = 1}^n (3i - 2) = $

Let $\alpha = 1^2 + 4^2 + 8^2 + 13^2 + 19^2 + 26^2 + \ldots$ up to $10$ terms and $\beta = \sum_{n=1}^{10} n^4$. If $4\alpha - \beta = 55k + 40$,then $k$ is equal to . . . . . . .

The sum of the series $3.6 + 4.7 + 5.8 + \dots$ up to $(n - 2)$ terms is:

The sum of $n$ terms of the series $1^{3}+3^{3}+5^{3}+7^{3}+\ldots$ is

The sum of all terms of the $n^{th}$ bracket of the sequence $(1), (3, 5), (7, 9, 11), \dots$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module