Prove that the coefficient of xn in the expansion of (1 + x)2n is twice the coefficient of xn in the expansion of (1 + x)2n – 1.

Asked by Abhisek | 1 year ago |  83

##### Solution :-

The general term Tr+1 in the binomial expansion is given by Tr+1 = nCr an-r br

The general term for binomial (1+x)2n is

Tr+1 = 2nCr xr …………………..1

To find the coefficient of xn

r = n

Tn+1 = 2nCn xn

The coefficient of xn = 2nCn

The general term for binomial (1+x)2n-1 is

Tr+1 = 2n-1Cr xr

To find the coefficient of xn

Putting n = r

Tr+1 = 2n-1Cr xn

The coefficient of xn = 2n-1Cn

We have to prove

Coefficient of xn in (1+x)2n = 2 coefficient of xn in (1+x)2n-1

Consider LHS = 2nCn

$$\dfrac{2!}{n!(2n-n)!}$$

$$\dfrac{2n!}{n!n!}$$

Consider RHS = $$2 \times ^{2n-1}C_n$$

$$2\times \dfrac{(2n-1)!}{n!(2n-1-n)!}$$

$$2\times \dfrac{(2n-1)!}{n!(n-1)!}$$

Now multiplying and dividing by n we get

$$2\times \dfrac{(2n-1)!}{n!(n-1)!}\times \dfrac{n}{n}$$

$$\dfrac{2n(2n-1)!}{n!n(n-1)!}$$

From above equations LHS = RHS

Hence the proof

Answered by Pragya Singh | 1 year ago

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