The nth term of the series 1+3+7+13+21+ldots | vedclass

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

Question

(A) Let the $n$th term of the series be $t_n$. The series is $1, 3, 7, 13, 21, \ldots$. Taking the differences between consecutive terms: $3-1=2, 7-3=

Vedclass · Accepted Answer

$100$

Vedclass · Answer

(A) Let the $n$th term of the series be $t_n$. The series is $1, 3, 7, 13, 21, \ldots$. Taking the differences between consecutive terms: $3-1=2, 7-3=4, 13-7=6, 21-13=8, \ldots$. This is an arithmetic progression of differences: $2, 4, 6, 8, \ldots$. The $n$th term $t_n$ can be expressed as $t_n = t_1 + \sum_{k=1}^{n-1} (2k)$. $t_n = 1 + 2 	imes \frac{(n-1)n}{2} = 1 + n(n-1) = n^2 - n + 1$. Given $t_n = 9901$,we have $n^2 - n + 1 = 9901$. $n^2 - n - 9900 = 0$. Factoring the quadratic equation: $n^2 - 100n + 99n - 9900 = 0$. $n(n-100) + 99(n-100) = 0$. $(n-100)(n+99) = 0$. Since $n$ must be a positive integer,$n = 100$.

The $n$th term of the series $1+3+7+13+21+\ldots$ is $9901$. The value of $n$ is

Similar Questions

If $S_n = 1^3 + 2^3 + \ldots + n^3$ and $T_n = 1 + 2 + \ldots + n$,then

Find the sum to $n$ terms of the series $3 \times 8 + 6 \times 11 + 9 \times 14 + \dots$

If $4^3+8^3+12^3+\ldots$ up to $n$ terms $= k n^2(n+1)^2$ (for all $n \in N$),then $k=$

The value of $\sum\limits_{r = 16}^{30} {(r + 2)(r - 3)}$ is equal to

Find the $20^{\text{th}}$ term of the series $2 \times 4 + 4 \times 6 + 6 \times 8 + \dots$ up to $n$ terms.

Vedclass Test Series

Exam Paper Generator

Online Exam Module