If the coefficient of x^7 in (ax^2+1bx)^11 equals the coefficient of x^-7 in (ax-1bx^2)^11…

Question

If the coefficient of $x^7$ in $\left(ax^2+\dfrac1{bx}\right)^{11}$ equals the coefficient of $x^{-7}$ in $\left(ax-\dfrac1{bx^2}\right)^{11}$, then $a$ and $b$ satisfy the relation:

Accepted Answer

General term of $${\left[ {a{x^2} + \left( {{1 \over {bx}}} \right)} \right]^{11}}$$ is Tr+1.

Tr+1 = $${}^{11}{C_r}{\left( {a{x^2}} \right)^{11 - r}}{\left( {{1 \over {bx}}} \right)^r}$$

= $${}^{11}{C_r}{\left( a \right)^{11 - r}}{\left( b \right)^{ - r}}{\left( x \right)^{22 - 3r}}$$

For the coefficient of x7,

$$ \Rightarrow $$ 22 - 3r = 7

$$ \Rightarrow $$ r = 5

So coefficient of x7 = $${}^{11}{C_5}{\left( a \right)^6}{\left( b \right)^{ - 5}}$$ ......(1)

Now General term of $${\left[ {ax - \left( {{1 \over {b{x^2}}}} \right)} \right]^{11}}$$ is Tr+1.

Tr+1 = $${}^{11}{C_r}{\left( {a{x}} \right)^{11 - r}}{\left( { - {1 \over {bx}}} \right)^r}$$

= $${}^{11}{C_r}{\left( a \right)^{11 - r}}{\left( { - 1} \right)^r}{\left( b \right)^{ - r}}{\left( x \right)^{11 - r}}{\left( x \right)^{ - 2r}}$$

For the coefficient of x-7,

11 - 3r = -7

$$ \Rightarrow $$ r = 6

$$\therefore$$ Coefficient of x-7 = $${}^{11}{C_6}{\left( a \right)^5}{\left( { - 1} \right)^6}{\left( b \right)^{ - 6}}$$

According to question,

Coefficient of x7 = Coefficient of x-7

$$ \Rightarrow $$ $${}^{11}{C_5}{\left( a \right)^6}{\left( b \right)^{ - 5}}$$ = $${}^{11}{C_6}{\left( a \right)^5}{\left( { - 1} \right)^6}{\left( b \right)^{ - 6}}$$

$$ \Rightarrow $$ $ab = 1$

Solutions · 1