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Prove that the coefficient of x^n in the expansion of (1+x)^{2

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Prove that the coefficient of $x^{n}$ in the expansion of $(1 + x)^{2 n}$ is twice the coefficient of $x^{n}$ in the expansion of $(1 + x)^{2 n - 1}$

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Prove that the coefficient of

x^{n}

in the expansion of

(1 + x)^{2 n}

is twice the coefficient of

x^{n}

in the expansion of

(1 + x)^{2 n - 1}

6 mins ago

Discuss this question LIVE

6 mins ago

Text solutionVerified

(1 + x)^{2 n} =^{2 n} C_{0} +^{2 n} C_{1} x^{1} +^{2 n} C_{2} x^{2} + ...... +^{2 n} C_{n} x^{n}

(1 + x)^{2 n - 1} =^{2 n - 1} C_{0} +^{2 n - 1} C_{1} x^{1} +^{2 n - 1} C_{2} x^{2} + ..... +^{2 n - 1} C_{n} x^{n}

Coefficient of

x^{n}

in (1 + x)^{2 n - 1} is (^{2 n - 1} C_{n})

^{2 n} C_{n} = \frac{( 2 n )!}{n ! n !}

= \frac{( 2 n ) ( 2 n - 1 ) ( 2 n - 2 )!}{n ( n - 1 )! n !}

= \frac{( 2 n ) ( 2 n - 1 ) 2 ( n - 1 )!}{n ( n - 1 )! n !}

= 2 [\frac{2 ( 2 n - 1 )!}{n !}] ............ (i)

Now,

^{2 n - 1} C_{n} = \frac{( 2 n - 1 )!}{n ! ( 2 n - 1 - n ) !}

= \frac{( 2 n - 1 )!}{n ! ( n - 1 ) !}

= \frac{( 2 n - 1 ) ( 2 n - 2 )!}{n ! ( n - 1 ) !}

= \frac{( 2 n - 1 ) 2 ( n - 1 )!}{n ! ( n - 1 ) !}

= \frac{2 ( 2 n - 1 )}{n !} ............. (ii)

From (i) and (ii), we have

^{2 n} C_{n} = 2 .^{2 n - 1} C_{n}

Hence, it is true.

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Question Text	Prove that the coefficient of $x^{n}$ in the expansion of $(1 + x)^{2 n}$ is twice the coefficient of $x^{n}$ in the expansion of $(1 + x)^{2 n - 1}$
Answer Type	Text solution:1
Upvotes	150