If the coefficients of x^7 in (x^2+1bx)^11 and x^-7 in (x-1bx^2)^11, b0, are equal, then t…

Question

If the coefficients of $x^7$ in $\left(x^2+\dfrac1{bx}ight)^{11}$ and $x^{-7}$ in $\left(x-\dfrac1{bx^2}ight)^{11}$, $b
e0$, are equal, then the value of $b$ is equal to

Accepted Answer

Coefficient of x7 in $${\left( {{x^2} + {1 \over {bx}}} \right)^{11}}$$ :

General Term = $${}^{11}{C_r}{({x^2})^{11 - r}}.{\left( {{1 \over {bx}}} \right)^r}$$

= $${}^{11}{C_r}{x^{22 - 3r}}.{1 \over {{b^r}}}$$

$22 - 3r = 7$

$r = 5$

$$\therefore$$ Required Term = $${}^{11}{C_5}.{1 \over {{b^5}}}.{x^7}$$

Coefficient of x-7 in $${\left( {x - {1 \over {b{x^2}}}} \right)^{11}}$$ :

General Term = $${}^{11}{C_r}{(x)^{11 - r}}.{\left( { - {1 \over {b{x^2}}}} \right)^r}$$

= $${}^{11}{C_r}{x^{11 - 3r}}.{{{{( - 1)}^r}} \over {{b^r}}}$$

$11 - 3r =  - 7$ $$\therefore$$ $r = 6$

$$ \therefore $$ Required Term = $${}^{11}{C_6}.{1 \over {{b^6}}}{x^{ - 7}}$$

According to the question,

$${}^{11}{C_5}.{1 \over {{b^5}}} = {}^{11}{C_6}.{1 \over {{b^6}}}$$

Since, b $$\ne$$ 0 $$\therefore$$ b = 1

Solutions · 1