Inthe expansion of (1+x)m+n where m and n are positive integers.show that coefficient of xm and xn are equal?
Tr+1 = nCran-rbr
Tr+1 = m+nCr(1)m+n-rxr
Comparing the indices of x in xm,
xr = xm
By putting the value of r, we get
=(m+n)! / (m+n-m)!m!
=(m+n)! / m!n! .........(1)
Comparing the indices of x in xn,
xr = xn
similarly, by putting the value of r we get,
Tn+1 = m+nCn(1)m+n-nxn
= (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.
The given expansion is (1 + x)m + n.
General term of the given expansion, tr = m + n Cr xr
Coefficient of xr in the expansion = m + n Cr
∴ Coefficient of x m in the given expansion = m + n Cm = ...(1)
Coefficient of xn in the given expansion = m + n Cn = ...(2)
From (1) and (2), we get
Coefficient of xm and xn in the given expansion are equal.
Thus, coefficient of x m in the given expansion is equal to coefficient of xn in the given expansion.