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

Binomial Theorem PYQ — JEE Main 2021 · 25 Jul · Shift 1 · Q29 with Solution

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Correct answer: C

Given sum

We need to approximate

S=\frac1{a-b}+\frac1{a-2b}+\frac1{a-3b}+\cdots+\frac1{a-nb} =\sum_{r=1}^n \frac1{a-rb}

under the condition that $b$ is very small compared to $a$ , so powers $(b/a)^3$ and higher can be neglected.

Rewrite each term for binomial expansion

\frac1{a-rb}=\frac1a\cdot \frac1{1-\frac{rb}{a}}

Now use

\frac1{1-x}=1+x+x^2+x^3+\cdots

Since terms involving $(b/a)^3$ and higher are neglected, we keep up to $x^2$ :

\frac1{1-x}\approx 1+x+x^2

with $x=\frac{rb}{a}$ .

Hence,

\frac1{a-rb}\approx \frac1a\left(1+\frac{rb}{a}+\frac{r^2b^2}{a^2}\right) =\frac1a+\frac{rb}{a^2}+\frac{r^2b^2}{a^3}

Sum over $r=1$ to $n$

So,

S\approx \sum_{r=1}^n \left(\frac1a+\frac{rb}{a^2}+\frac{r^2b^2}{a^3}\right)

S\approx \frac1a\sum_{r=1}^n 1+\frac{b}{a^2}\sum_{r=1}^n r+\frac{b^2}{a^3}\sum_{r=1}^n r^2

Using standard formulas,

\sum_{r=1}^n 1=n, \qquad \sum_{r=1}^n r=\frac{n(n+1)}2, \qquad \sum_{r=1}^n r^2=\frac{n(n+1)(2n+1)}6

Thus,

S\approx \frac{n}{a}+\frac{b}{a^2}\cdot \frac{n(n+1)}2+\frac{b^2}{a^3}\cdot \frac{n(n+1)(2n+1)}6

Extract the coefficient of $n^3$

We are given

S=\alpha n+\beta n^2+\gamma n^3

So we only need the coefficient of $n^3$ .

Now,

\frac{n(n+1)(2n+1)}6=\frac{2n^3+3n^2+n}{6}

Hence the $n^3$ contribution comes only from

\frac{b^2}{a^3}\cdot \frac{2n^3+3n^2+n}{6}

So,

\gamma=\frac{b^2}{a^3}\cdot \frac{2}{6}=\frac{b^2}{3a^3}

Match with options

\gamma=\frac{b^2}{3a^3}

This corresponds to Option C.

Comparison with stored answer

Stored correct answer: C

Our derived answer: C

So they agree.

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View written solutionFree

Correct answer: C

Given sum

We need to approximate

S=\frac1{a-b}+\frac1{a-2b}+\frac1{a-3b}+\cdots+\frac1{a-nb} =\sum_{r=1}^n \frac1{a-rb}

under the condition that $b$ is very small compared to $a$ , so powers $(b/a)^3$ and higher can be neglected.

Rewrite each term for binomial expansion

\frac1{a-rb}=\frac1a\cdot \frac1{1-\frac{rb}{a}}

Now use

\frac1{1-x}=1+x+x^2+x^3+\cdots

Since terms involving $(b/a)^3$ and higher are neglected, we keep up to $x^2$ :

\frac1{1-x}\approx 1+x+x^2

with $x=\frac{rb}{a}$ .

Hence,

\frac1{a-rb}\approx \frac1a\left(1+\frac{rb}{a}+\frac{r^2b^2}{a^2}\right) =\frac1a+\frac{rb}{a^2}+\frac{r^2b^2}{a^3}

Sum over $r=1$ to $n$

So,

S\approx \sum_{r=1}^n \left(\frac1a+\frac{rb}{a^2}+\frac{r^2b^2}{a^3}\right)

S\approx \frac1a\sum_{r=1}^n 1+\frac{b}{a^2}\sum_{r=1}^n r+\frac{b^2}{a^3}\sum_{r=1}^n r^2

Using standard formulas,

\sum_{r=1}^n 1=n, \qquad \sum_{r=1}^n r=\frac{n(n+1)}2, \qquad \sum_{r=1}^n r^2=\frac{n(n+1)(2n+1)}6

Thus,

S\approx \frac{n}{a}+\frac{b}{a^2}\cdot \frac{n(n+1)}2+\frac{b^2}{a^3}\cdot \frac{n(n+1)(2n+1)}6

Extract the coefficient of $n^3$

We are given

S=\alpha n+\beta n^2+\gamma n^3

So we only need the coefficient of $n^3$ .

Now,

\frac{n(n+1)(2n+1)}6=\frac{2n^3+3n^2+n}{6}

Hence the $n^3$ contribution comes only from

\frac{b^2}{a^3}\cdot \frac{2n^3+3n^2+n}{6}

So,

\gamma=\frac{b^2}{a^3}\cdot \frac{2}{6}=\frac{b^2}{3a^3}

Match with options

\gamma=\frac{b^2}{3a^3}

This corresponds to Option C.

Comparison with stored answer

Stored correct answer: C

Our derived answer: C

So they agree.

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