First slide

Binomial theorem for positive integral Index

Question

If the coefficient of x⁷ in ${[{ax}^{2} + (\frac{1}{bx})]}^{11}$ equals the coefficient of $x^{- 7}$ in ${[ax - (\frac{1}{{bx}^{2}})]}^{11}$ , then a and b satisfy the relation

Moderate

A

a + b = 1
B

a - b = 1
C

ab = 1
D

$\frac{a}{b} = 1$

Solution

$\begin{array}{l} For {({ax}^{2} + (\frac{1}{bx}))}^{11}, \\ T_{r + 1} =^{11} C_{r} {({ax}^{2})}^{11 - r} {(\frac{1}{bx})}^{r} =^{11} C_{r} a^{11 - r} \frac{1}{b^{r}} x^{22 - 3 r} \\ For x^{7}, \\ 22 - 3 r = 7 \\ or 3 r = 15 \\ or r = 5 \\ \Rightarrow T_{6} =^{11} C_{5} a^{6} \frac{1}{b^{5}} x^{7} \\ \Rightarrow Coefficient of x^{7} {is}^{11} C_{5} \frac{a^{6}}{b^{5}} \end{array}$
similarly, coefficient of x^-7 in ${(ax - (\frac{1}{{bx}^{2}}))}^{11} {is}^{11} C_{6} \frac{a^{5}}{b^{6}}$ .
Given that $^{11} C_{5} \frac{a^{6}}{b^{5}} =^{11} C_{6} \frac{a^{5}}{b_{6}}$
$\Rightarrow a = \frac{1}{b}$
or ab = 1

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