"If the coefficients of x7 in ${\left( {{x^2} + {1 \over {bx}}} \right)^{11}}$ and x$-$7 in ${\left( {{x} - {1 \over {bx^2}}} \right)^{11}}$, b $\ne$ 0, are equal, then the value of b is equal to :"

Question

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"

If the coefficients of x⁷ in ${\left( {{x^2} + {1 \over {bx}}} \right)^{11}}$ and x^$-$7 in ${\left( {{x} - {1 \over {bx^2}}} \right)^{11}}$, b $\ne$ 0, are equal, then the value of b is equal to :

Solution

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If the coefficients of x7 in ${\left( {{x^2} + {1 \over {bx}}} \right)^{11}}$ and x$-$7 in ${\left( {{x} - {1 \over {bx^2}}} \right)^{11}}$, b $\ne$ 0, are equal, then the value of b is equal to :

Solution

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More Binomial Theorem questions

Drill 25 more like these. Every day. Free.

If the coefficients of x⁷ in ${\left( {{x^2} + {1 \over {bx}}} \right)^{11}}$ and x^$-$7 in ${\left( {{x} - {1 \over {bx^2}}} \right)^{11}}$, b $\ne$ 0, are equal, then the value of b is equal to :