Inthe expansion of (1+x)m+n where m and n are positive integers show that coefficient of xm and xn are equal - Maths - Binomial Theorem

Question

Inthe expansion of (1+x)m+n where m and n are
positive integers show that coefficient of xm and
xn are equal?

Dharu Shingadia · Accepted Answer

(1+x)m+n
Tr+1 =
nCran-rbr
Tr+1 =
m+nCr(1)m+n-rxr
Comparing the indices of x in xm,
xr = xm
=> r=m
By putting the value of r, we get
Tm+1=
m+nCm(1)m+n-mxm
= m+ncmxm
Coefficients= m+ncm
=(m+n)! / (m+n-m)!m!
=(m+n)! / m!n! (1)
Comparing the indices of x in xn,
xr = xn
=> r=n
similarly, by putting the value of r we get,
Tn+1 =
m+nCn(1)m+n-nxn
Coefficient= m+nCn
= (m+n)! / (m+n-n)!n!
=(m+n)! / m!n! (2)
From eq (1) & (2), we can say that coefficients of
xm & xn are equal

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