JEE Mains · Maths · STD 11 - 7. binomial theoram
If \(\mathrm{b}\) is very small as compared to the value of \(\mathrm{a}\), so that the cube and other higher powers of \(\frac{b}{a}\) can be neglected in the identity \(\frac{1}{a-b}+\frac{1}{a-2 b}+\frac{1}{a-3 b} \ldots .+\frac{1}{a-n b}=\alpha n+\beta n^{2}+\gamma n^{3}\), then the value of \(\gamma\) is:
- A \(\frac{b^{2}}{3 a^{3}}\)
- B \(\frac{a+b}{3 a^{2}}\)
- C \(\frac{a^{2}+b}{3 a^{3}}\)
- D \(\frac{a+b^{2}}{3 a^{3}}\)
Answer & Solution
Correct Answer
(A) \(\frac{b^{2}}{3 a^{3}}\)
Step-by-step Solution
Detailed explanation
\((a-b)^{-1}+(a-2 b)^{-1}+\ldots \ldots+(a-n b)^{1}\) \(=\frac{1}{a} \sum_{r=1}^{n}\left(1-\frac{r b}{a}\right)^{-1}\) \(=\frac{1}{a} \sum_{r=1}^{n}\left\{\left(1+\frac{r b}{a}+\frac{r^{2} b^{2}}{a^{2}}\right)+(\right.\) terms to be neglected \(\left.)\right\}\)…
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