Question

If 'a''a'  and 'b''b'  are the respective coefficients of 'xm''xm'  and 'xn''xn'  in the expansion of (1+x)m+n,(1+x)m+n,  then​

Answer 1

ANSWER

Using Binomial theorem,

(1+x)m+n=k=0∑m+n(m+nk)xk

Tk+1=(m+nk)xk

∴ Coefficient of xm,a=(m+nm)=m!(m+n−m)!(m+n)!

a=m!n!(m+n)!

Coefficient of xn,b=(m+nn)=n!(m+n−n)!(m+n)!

b=n!m!m+n)!

∴a+b=m!n!(m+n)!+m!n!(m+n)!

⇒a+b=2a

⇒b=2a−a

⇒b=a.

Answer 2

Answer:

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