Find the coefficient of x 7 in the expansion of a x 2 - 1 bx 11

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Standard XII

Mathematics

Binomial Coefficients

Find the coef...

Question

Find the coefficient of $x^{7}$ in the expansion of ${(a x^{2} + \frac{1}{b x})}^{11}$

{^{11} C}_{6} a^{6} b^{- 5}

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{^{11} C}_{5} a^{6} b^{- 6}

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{^{11} C}_{5} a^{7} b^{- 5}

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{^{11} C}_{5} a^{6} b^{- 5}

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Solution

The correct option is C ${^{11} C}_{5} a^{6} b^{- 5}$
The general term for the given expression will be
$T_{r + 1} =^{11} C_{r} a^{11 - r} . b^{- r} x^{22 - 3 r}$
Therefore, for $x^{7}$
$22 - 3 r = 7$
$15 = 3 r$
$r = 5$ .
Hence
$T_{6} =^{11} C_{5} a^{6} b^{- 5} . x^{7}$
Therefore the coefficient is
$^{11} C_{5} a^{6} b^{- 5}$ .

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Similar questions

Suppose $a_{2}, a_{3}, a_{4}, a_{5}, a_{6}, a_{7}$ are integers such that $\frac{5}{7} = \frac{a_{2}}{2!} + \frac{a_{3}}{3!} + \frac{a_{4}}{4!} + \frac{a_{5}}{5!} + \frac{a_{6}}{6!} + \frac{a_{7}}{7!}$ where 0 $\leq$ a< j for j= 2,4,5,6,7.

The sum $a_{2} + a_{3} + a_{4} + a_{5} + a_{6} + a_{7}$ is

Find the coefficient of $x^{11}$ in the expansion $(1 - 6 x^{6}) (1 - x)^{- 7}$

Q. Suppose

a_{2}, a_{3}, a_{4}, a_{5}, a_{6}, a_{7}

are integers such that

\frac{5}{7} = \frac{a_{2}}{2!} + \frac{a_{3}}{3!} + \frac{a_{4}}{4} + \frac{a_{5}}{5!} + \frac{a_{6}}{6!} + \frac{a_{7}}{7!}

where

0 \leq a < j

for

j = 2, 4, 5, 6, 7.

The sum

a_{2} + a_{3} + a_{4} + a_{5} + a_{6} + a_{7}

Q. Suppose

a_{2}, a_{3}, a_{4}, a_{5}, a_{6}, a_{7}

are intagers such that

\frac{5}{7} = \frac{a_{2}}{2!} + \frac{a_{3}}{3!} + \frac{a_{4}}{4!} + \frac{a_{5}}{5!} + \frac{a_{6}}{6!} + \frac{a_{7}}{7!}

where

0 \leq a_{j} < j

for

j = 2, 3, 4, 5, 6, 7

. The sum

a_{2} + a_{3} + a_{4} + a_{5} + a_{6} + a_{7}

is-

Q. If the coefficient of

x^{7}

{[a x^{2} + (\frac{1}{b x})]}^{11}

equals the coefficient of

x^{- 7}

{[a x^{2} - (\frac{1}{b x})]}^{11}

, then

a

and

b

satisfy the relation

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Find the coefficient of x7 in the expansion of (ax2+1bx)11

The correct option is C 11C5a6b−5The general term for the given expression will be Tr+1=11Cra11−r.b−rx22−3rTherefore, for x722−3r=715=3rr=5.Hence T6=11C5a6b−5.x7Therefore the coefficient is11C5a6b−5.

Find the coefficient of $x^{7}$ in the expansion of ${(a x^{2} + \frac{1}{b x})}^{11}$