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JEE Main 202131 AugustEvening Shift
Exam
JEE Main
Year
2021
Date
31 August
Shift
Evening Shift
Subject
Mathematics
Chapter
Binomial Theorem

JEE Main 2021 · 31 August · Evening Shift · Mathematics · Binomial Theorem

If the coefficient of a ^ 7 b ^ 8 in the expansion of ( a + 2b +...

If the coefficient of a7b8{a}^{7}{b}^{8} in the expansion of (a+2b+4ab)10( a + 2b + 4ab{)}^{10} is K⋅216,K \cdot {2}^{16}, then KK is equal to

Correct answer

0

Step-by-step solution

Given the expansion(a+2b+4ab)10( a + 2b + 4ab{)}^{10}

Tr+1=a10⋅b10(1b+2a+4)10{T}_{r + 1} = {a}^{10} \cdot {b}^{10}{\left(\dfrac{1}{b} + \dfrac{2}{a} + 4\right)}^{10}

Tr+1=a10⋅b10C10r1+r2(1b)r1(2a)r2410−r1−r2{T}_{r + 1} = {a}^{10} \cdot {b}^{10}{C 10}_{{r}_{1} + {r}_{2}}{\left(\dfrac{1}{b}\right)}^{{r}_{1}}{\left(\dfrac{2}{a}\right)}^{{r}_{2}}{4}^{10 - {r}_{1} - {r}_{2}}

Tr+1=a10⋅b1010!(1b)r1(2a)r2410−r1−r2r1!r2!(10−r1−r2)!{T}_{r + 1} = {a}^{10} \cdot {b}^{10}\dfrac{10 ! {\left(\dfrac{1}{b}\right)}^{{r}_{1}}{\left(\dfrac{2}{a}\right)}^{{r}_{2}}{4}^{10 - {r}_{1} - {r}_{2}}}{{r}_{1} ! {r}_{2} ! \left(10 - {r}_{1} - {r}_{2}\right) !}

Putting, r1=2,r2=3{r}_{1} = 2, {r}_{2} = 3

Coefficient of a7b8{a}^{7}{b}^{8} is 10!⋅23⋅410−2−32!.3!5!\dfrac{10 ! \cdot {2}^{3} \cdot {4}^{10 - 2 - 3}}{2 ! . 3 ! 5 !}

=213⋅10!2!.3!5!= \dfrac{{2}^{13} \cdot 10 !}{2 ! . 3 ! 5 !}

=216⋅315= {2}^{16} \cdot 315

K.216=315..216K . {2}^{16} = 315 . . {2}^{16}

Hence, K=315K = 315

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