The n^{text{th}} term of the series 1 + (3 + | vedclass

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

Question

(D) The given series is $1 + (3 + 5 + 7) + (9 + 11 + 13 + 15 + 17) + \ldots$ The number of terms in the $n^{	ext{th}}$ group is $2n - 1$. Let the fir

Vedclass · Accepted Answer

$(2n - 1)[(n - 1)^2 + n^2]$

Vedclass · Answer

(D) The given series is $1 + (3 + 5 + 7) + (9 + 11 + 13 + 15 + 17) + \ldots$ The number of terms in the $n^{	ext{th}}$ group is $2n - 1$. Let the first term of the $n^{	ext{th}}$ group be $a_n$. The total number of terms before the $n^{	ext{th}}$ group is $1 + 3 + 5 + \ldots + (2(n-1) - 1) = (n-1)^2$. Thus,the first term of the $n^{	ext{th}}$ group is $a_n = (n-1)^2 + 1$. The $n^{	ext{th}}$ group is an arithmetic progression with $2n-1$ terms,first term $a_n$,and common difference $d = 2$. The sum of the $n^{	ext{th}}$ group $t_n$ is: $t_n = \frac{2n-1}{2} [2a_n + (2n-2)d] = (2n-1) [a_n + (n-1)2]$ $t_n = (2n-1) [(n-1)^2 + 1 + 2n - 2] = (2n-1) [n^2 - 2n + 1 + 1 + 2n - 2] = (2n-1) [n^2 + (n-1)^2]$.

The $n^{\text{th}}$ term of the series $1 + (3 + 5 + 7) + (9 + 11 + 13 + 15 + 17) + \ldots$ is:

Similar Questions

$1 + \frac{1^3 + 2^3}{1 + 2} + \frac{1^3 + 2^3 + 3^3}{1 + 2 + 3} + \dots + \frac{1^3 + 2^3 + 3^3 + \dots + 15^3}{1 + 2 + 3 + \dots + 15} - \frac{1}{2}(1 + 2 + 3 + \dots + 15)$ is equal to

If the sum of the first ten terms of the series ${\left( {1\frac{3}{5}} \right)^2} + {\left( {2\frac{2}{5}} \right)^2} + {\left( {3\frac{1}{5}} \right)^2} + {4^2} + \dots$ is $\frac{16}{5}m$,then $m$ is equal to:

Write the first five terms of the sequence whose $n^{th}$ term is $a_{n} = n \frac{n^{2}+5}{4}$.

The minimum value of $n$ for which $\frac{2^2+4^2+6^2+\ldots+(2n)^2}{1^2+3^2+5^2+\ldots+(2n-1)^2} < 1.01$ 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:

Vedclass Test Series

Exam Paper Generator

Online Exam Module