The coefficient of x7 and x8 in the expansion of (2+3x ) n are equal, then n is equal to-

[2+3x ] n(2)n[1+6x ] n[1+6x ] ncoefficient of x7= nC7(6x )7 = nC7(61) 7(x) 7coefficient of x8= nC8(6x ) 8= nC8(61 ) 8(x) 8given(2)n nC7(61 ) 7=(2)n nC8(61 ) 8 nC7=61 nC8(n−7)!7!n!=61(n−8)!8!n!8×6(n−7)!7!=(n−7)(n−8)!7!48=n−7⇒n=48+7=55⇒n=55

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The coefficient of {x}^{7} and {x}^{8} in the expansion of { (

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The coefficient of $x^{7}$ and $x^{8}$ in the expansion of $(2 + \frac{x}{3})^{n}$ are equal, then $n$ is equal to-

35

45

55

N o n e o f t h ese

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The coefficient of

x^{7}

and

x^{8}

in the expansion of

(2 + \frac{x}{3})^{n}

are equal, then

n

is equal to-

6 mins ago

Discuss this question LIVE

6 mins ago

Text solutionVerified

[2 + \frac{x}{3}]^{n}

(2)^{n} [1 + \frac{x}{6}]^{n}

[1 + \frac{x}{6}]^{n}

coefficient of

x^{7} =_{n} C_{7} (\frac{x}{6})^{7} =_{n} C_{7} (\frac{1}{6})^{7} (x)^{7}

coefficient of

x^{8} =_{n} C_{8} (\frac{x}{6})^{8} =_{n} C_{8} (\frac{1}{6})^{8} (x)^{8}

given

(2)^{n}_{n} C_{7} (\frac{1}{6})^{7} = (2)^{n}_{n} C_{8} (\frac{1}{6})^{8}

_{n} C_{7} = \frac{1}{6}_{n} C_{8}

\frac{n !}{( n - 7 ) ! 7 !} = \frac{1}{6} \frac{n !}{( n - 8 ) ! 8 !}

8 \times 6 (n - 7)! 7! = (n - 7) (n - 8)! 7!

48 = n - 7 \Rightarrow n = 48 + 7 = 55 \Rightarrow n = 55

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Question Text	The coefficient of $x^{7}$ and $x^{8}$ in the expansion of $(2 + \frac{x}{3})^{n}$ are equal, then $n$ is equal to-
Answer Type	Text solution:1
Upvotes	150

The coefficient of x7 and x8 in the expansion of (2+3x​ ) n are equal, then n is equal to-

Text solutionVerified

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The coefficient of $x^{7}$ and $x^{8}$ in the expansion of $(2 + \frac{x}{3})^{n}$ are equal, then $n$ is equal to-