If the coefficients of x7 and x8 in the expansion of 2-x-3n are equal then n-A- 55B- 35C- 27D- 45

The correct option is A 55 ( 2+x/3 )^n =2^n ( 1+x/6 )^n Coefficient of x^7 is 2^n . ^nC_7/6^7. Coefficient of x^8 is 2^n .^nC_8/6^8 Coefficient of x^7 = co ...

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Byju's Answer

Standard XII

Mathematics

Sum of Product of Binomial Coefficients

If the coeffi...

Question

If the coefficients of $x^{7} a n d x^{8}$ in the expansion of ${(2 + \frac{x}{3})}^{n}$ are equal then n =

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Solution

The correct option is B
55

${(2 + \frac{x}{3})}^{n} = 2^{n} {(1 + \frac{x}{6})}^{n}$

$C o e f f i c i e n t o f x^{7} i s \frac{2^{n} .^{n} C_{7}}{6^{7}} . C o e f f i c i e n t o f x^{8} i s \frac{2^{n} .^{n} C_{8}}{6^{8}}$

$C o e f f i c i e n t o f x^{7} = c o e f f i c i e n t o f x^{8} \Rightarrow \frac{2^{n} .^{n} C_{7}}{6^{7}} = 2 n . \frac{^{n} C_{8}}{6^{8}} \Rightarrow 6 (^{n} C_{7}) =^{n} C_{8}$

$\Rightarrow 6. \frac{n!}{(n - 7)! 7!} = \frac{n!}{(n - 8)! 8!} \Rightarrow \frac{6}{n - 7} = \frac{1}{8} \Rightarrow n - 7 = 48 \Rightarrow n = 55$

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If the coefficients of x7 and x8 in the expansion of (2+x3)n are equal then n =

The correct option is B 55 (2+x3)n=2n(1+x6)n Coefficient of x7 is2n.nC767. Coefficient of x8 is 2n.nC868 Coefficient of x7=coefficient of x8⇒2n.nC767=2n.nC868⇒6(nC7)=nC8 ⇒6.n!(n−7)!7!=n!(n−8)!8!⇒6n−7=18⇒n−7=48⇒n=55

If the coefficients of $x^{7} a n d x^{8}$ in the expansion of ${(2 + \frac{x}{3})}^{n}$ are equal then n =