If b is very small as compared to the value o | vedclass

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

Question

(A) The given expression is $S = \sum_{r=1}^{n} \frac{1}{a-rb} = \frac{1}{a} \sum_{r=1}^{n} (1 - \frac{rb}{a})^{-1}$.Using the binomial expansion $(1-

Vedclass · Accepted Answer

$\frac{b^2}{3a^3}$

Vedclass · Answer

(A) The given expression is $S = \sum_{r=1}^{n} \frac{1}{a-rb} = \frac{1}{a} \sum_{r=1}^{n} (1 - \frac{rb}{a})^{-1}$.Using the binomial expansion $(1-x)^{-1} = 1 + x + x^2 + \dots$,we have:$S = \frac{1}{a} \sum_{r=1}^{n} (1 + \frac{rb}{a} + \frac{r^2b^2}{a^2} + \dots)$.Neglecting higher powers of $\frac{b}{a}$,we get:$S = \frac{1}{a} [n + \frac{b}{a} \sum r + \frac{b^2}{a^2} \sum r^2]$.Using sum formulas $\sum_{r=1}^{n} r = \frac{n(n+1)}{2}$ and $\sum_{r=1}^{n} r^2 = \frac{n(n+1)(2n+1)}{6} \approx \frac{n^3}{3}$ for large $n$ or focusing on the $n^3$ coefficient:$S = \frac{n}{a} + \frac{b}{a^2} \frac{n^2}{2} + \frac{b^2}{a^3} \frac{2n^3}{6} + \dots = \frac{n}{a} + \frac{b}{2a^2} n^2 + \frac{b^2}{3a^3} n^3$.Comparing with $\alpha n + \beta n^2 + \gamma n^3$,we get $\gamma = \frac{b^2}{3a^3}$.

If $b$ is very small as compared to the value of $a$,so that the cube and other higher powers of $\frac{b}{a}$ can be neglected in the identity $\frac{1}{a-b}+\frac{1}{a-2b}+\frac{1}{a-3b}+\ldots+\frac{1}{a-nb}=\alpha n+\beta n^2+\gamma n^3$,then the value of $\gamma$ is:

Similar Questions

If $C_{j}$ stands for ${ }^{n} C_{j}$,then $\frac{C_0}{2} + \frac{C_1}{2 \cdot 2^2} + \frac{C_2}{3 \cdot 2^3} + \ldots + \frac{C_{n}}{(n+1) 2^{n+1}} = $

Find the coefficient of $x^{5}$ in the product $(1+2x)^{6}(1-x)^{7}$ using the binomial theorem.

The value of $(\sqrt{2} + 1)^6 + (\sqrt{2} - 1)^6$ is

The coefficient of $x^{48}$ in $(1+x)+2(1+x)^2+3(1+x)^3+ . . . +100(1+x)^{100}$ is equal to:

The expression $(2 + \sqrt{2})^4$ has a value lying between

Vedclass Test Series

Exam Paper Generator

Online Exam Module