Q11.    Prove that the coefficient of x^n in the expansion of (1+x)^{2n} is twice the coefficient of x^n in the expansion of (1+x)^{2n-1}.

Answers (1)

As we know that the general  (r+1)^{th} term  T_{r+1} in the binomial expansion of  (a+b)^n  is given by 

T_{r+1}=^nC_ra^{n-r}b^r

So, general  (r+1)^{th} term  T_{r+1} in the binomial expansion of  (1+x)^{2n} is,

T_{r+1}=^{2n}C_r1^{2n-r}x^r

x^n will come when r=n,

So, Coefficient of x^n in the binomial expansion of  (1+x)^{2n} is,

K_{1x^n}=^{2n}C_n

Now,

the general  (r+1)^{th} term  T_{r+1} in the binomial expansion of  (1+x)^{2n-1} is,

T_{r+1}=^{2n-1}C_r1^{2n-1-r}x^r

Here also x^n will come when r=n,

So, Coefficient of x^n in the binomial expansion of  (1+x)^{2n-1} is,

K_{2x^n}=^{2n-1}C_n

Now, As we can see

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

^{2n-1}C_n=\frac{1}{2}\times^{2n}C_n

2\times^{2n-1}C_n=^{2n}C_n

2\times K_{2x^n}=K_{1x^n}

Hence, the coefficient of x^n in the expansion of (1+x)^{2n} is twice the coefficient of x^n in the expansion of (1+x)^{2n-1}.

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