For a series whose n^{th} term is left( frac{ | vedclass

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

Question

(A) The $n^{th}$ term is given by $T_n = \frac{n}{x} + y$.To find the sum of $r$ terms,we calculate the sum $S_r = \sum_{n=1}^{r} T_n = \sum_{n=1}^{r}

Vedclass · Accepted Answer

$\left\{ \frac{r(r + 1)}{2x} \right\} + ry$

Vedclass · Answer

(A) The $n^{th}$ term is given by $T_n = \frac{n}{x} + y$.To find the sum of $r$ terms,we calculate the sum $S_r = \sum_{n=1}^{r} T_n = \sum_{n=1}^{r} \left( \frac{n}{x} + y \right)$.$S_r = \frac{1}{x} \sum_{n=1}^{r} n + \sum_{n=1}^{r} y$.Using the formula for the sum of the first $r$ natural numbers,$\sum_{n=1}^{r} n = \frac{r(r + 1)}{2}$.$S_r = \frac{1}{x} \left( \frac{r(r + 1)}{2} \right) + ry$.$S_r = \left\{ \frac{r(r + 1)}{2x} \right\} + ry$.

For a series whose $n^{th}$ term is $\left( \frac{n}{x} \right) + y$,the sum of $r$ terms is:

Similar Questions

The value of the expression $1.(2 - \omega )(2 - {\omega ^2}) + 2.(3 - \omega )(3 - {\omega ^2}) + ....... + (n - 1).(n - \omega )(n - {\omega ^2}),$ where $\omega$ is an imaginary cube root of unity,is

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

The sum of the series $1 + 3x + 6x^2 + 10x^3 + \dots \infty$ is

If the sum of the first $n$ terms of a series is $5n^2 + 2n$,then its second term is

The sum of the first $n$ terms of the series $\frac{1}{2} + \frac{3}{4} + \frac{7}{8} + \frac{15}{16} + \dots$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module