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Binomial theorem for positive integral Index

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Question

If the coefficients x⁷ in ${({ax}^{2} + \frac{1}{bx})}^{11}$ and coefficient of x^-7in ${(ax - \frac{1}{{bx}^{2}})}^{11}$ are equal, then the value of ab is _____.

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Solution

Let $x^{7} occurs in T_{r + 1}$ term, then
$T_{r + 1} =^{n} C_{r} {({ax}^{2})}^{n - r} {(\frac{1}{bx})}^{r} =^{11} C_{r} \frac{a^{11 - r}}{b^{r}} \cdot x^{22 - 2 r - r}$
For $x^{7}, 22 - 3 r = 7 or r = 5$
Hence, coefficients of $x^{7} {is}^{11} C_{5} \frac{a^{6}}{b^{5}}$ .
Let $x^{- 7} occur in T_{r + 1}$ term then
$T_{r + 1} =^{11} C_{r} (ax)^{11 - {(- \frac{1}{{bx}^{2}})}^{r}} =^{11} C_{r} \frac{a^{11 - r}}{(- b)^{r}} x^{11 - 3 r}$
For $x^{- 7} \Rightarrow 11 - 3 r = - 7 \Rightarrow r = 6$
Hence, coefficient of $x^{- 7} {is}^{11} C_{6} \frac{a^{5}}{b^{6}}$
Now, $^{11} C_{5} \frac{a^{5}}{b^{6}} =^{11} C_{6} \frac{a^{6}}{b^{5}}$
$\begin{array}{l} \Rightarrow^{11} C_{5} a =^{11} C_{6} \frac{a^{5}}{b^{6}} \\ \Rightarrow^{11} C_{5} a =^{11} C_{11 - 6} \frac{1}{b} \\ \Rightarrow^{11} C_{5} a =^{11} C_{5} \frac{1}{b} \end{array} \Rightarrow ab = 1$

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