Question:
The coefficient of $x^{-7}$ in the expansion of $\left[ax - \frac{1}{bx^{2}}\right]^{11} $ will be
The coefficient of $x^{-7}$ in the expansion of $\left[ax - \frac{1}{bx^{2}}\right]^{11} $ will be
Updated On: Jul 7, 2022
- $\frac{462}{b^5} a^6$
- $\frac{462 a^5}{b^6} $
- $\frac{ - 462 a^5}{b^6} $
- $\frac{ - 462 a^6}{b^5} $
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The Correct Option is B
Solution and Explanation
Suppose $x^{-7}$ occurs in $(r + 1)^{th}$ term.
we have $T_{r + 1} = {^{n}C_{r}} x^{n - r} \, a^r$ in $(x + a)^n$.
In the given question, n = 1, x = ax, a = $\frac{- 1}{bx^2} $
$\therefore \, T_{r + 1} = {^{11}C_{r}} (ax)^{11 - r} \left( \frac{ - 1}{bx^2} \right)^r$
$ = {^{11}C_r} a^{11 - r} b^{-r} x^{11 - 3r} (-1)^r$
This term contains $x^{-7} $ if $11 - 3r = - 7$
$\Rightarrow \, r = 6$
Therefore, coefficient of $x^{-7}$ is
${^{11}C_6 } (a)^5 \left( \frac{ -1}{b} \right)^6 = \frac{462}{b^6} a^5$
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Notes on binomial expansion formula
Concepts Used:
Binomial Expansion Formula
The binomial expansion formula involves binomial coefficients which are of the form
(n/k)(or) nCk and it is calculated using the formula, nCk =n! / [(n - k)! k!]. The binomial expansion formula is also known as the binomial theorem. Here are the binomial expansion formulas.
This binomial expansion formula gives the expansion of (x + y)n where 'n' is a natural number. The expansion of (x + y)n has (n + 1) terms. This formula says:
We have (x + y)n = nC0 xn + nC1 xn-1 . y + nC2 xn-2 . y2 + … + nCn yn
General Term = Tr+1 = nCr xn-r . yr
- General Term in (1 + x)n is nCr xr
- In the binomial expansion of (x + y)n , the rth term from end is (n – r + 2)th .