If the coefficient of x 7 in ( a x 2 + 1 b x ) 11 is equal to the coefficient of x − 7 in ( a x − 1 b x 2 ) 11 , then ab = Options: A . &nbsp 1 B . &nbsp 1 2 C . &nbsp 2 D . &nbsp 3

Answer: Option A : A In the expansion of ( a x 2 + 1 b x ) 11 , the general term is T r + 1 = 11 C r ( a x 2 ) 11 − r ( 1 b x ) r = 11 C r ( a ) 11 − r ( 1 b r ) x 22 − 3 r For x 7 , we must have 22 - 3r = 7 ⇒ r = 5, and the coefficient of x 7 = 11 C 5 . a 11 − 5 1 5 ) = 11 C 5 a 6 b 5 Similarly, in the expansion of ( a x − 1 b x 2 ) 11 , the general term is T r + 1 = 11 C r ( − 1 ) r a 11 − r b r . x 11 − 3 r For x − 7 we must have, 11 - 3r = -7 ⇒ r = 6, and the coefficient of x − 7 is 11 C 6 a 5 b 6 = 11 C 5 a 5 b 6 . As given, 11 C 5 a 6 b 5 = 11 C 5 a 5 b 6 ⇒ ab = 1.

If the coefficient of x7 in (ax2+1bx)11 is equal to the coefficient of x

Discussion Forum : Binomial Theorem

Question -

If the coefficient of $x^{7}$ in $(a x^{2} + \frac{1}{b x})^{11}$ is equal to the coefficient of $x^{- 7}$ in $(a x - \frac{1}{b x^{2}})^{11}$ , then ab =

Options:

A . 1

B .

\frac{1}{2}

C . 2

D . 3

Answer: Option A
:
A

In the expansion of $(a x^{2} + \frac{1}{b x})^{11}$ , the general

term is

$T_{r + 1}$ = $^{11} C_{r}$ $(a x^{2})^{11 - r}$ $(\frac{1}{b x})^{r}$ = $^{11} C_{r}$ $(a)^{11 - r}$ $(\frac{1}{b^{r}}) x^{22 - 3 r}$

For $x^{7}$ , we must have 22 - 3r = 7 ⇒ r = 5, and

the coefficient of $x^{7}$ = $^{11} C_{5}$ . $a^{11 - 5} \frac{1}{5}$ ) = $^{11} C_{5}$ $\frac{a^{6}}{b^{5}}$

Similarly, in the expansion of $(a x - \frac{1}{b x^{2}})^{11}$ , the

general term is $T_{r + 1}$ = $^{11} C_{r} (- 1)^{r}$ $\frac{a^{11 - r}}{b^{r}}$ . $x^{11 - 3 r}$

For $x^{- 7}$ we must have, 11 - 3r = -7 ⇒ r = 6, and

the coefficient of $x^{- 7}$ is $^{11} C_{6}$ $\frac{a^{5}}{b^{6}}$ = $^{11} C_{5}$ $\frac{a^{5}}{b^{6}}$ .

As given, $^{11} C_{5}$ $\frac{a^{6}}{b^{5}}$ = $^{11} C_{5}$ $\frac{a^{5}}{b^{6}}$ ⇒ ab = 1.

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