PURE MATHS BINOMIAL一問? 急!!!!

Consider the identity:
(1+x)^n+(1+x)^(n+1)+......+(1+x)^(n+m)=[(1+x)^(n+m+1)-(1+x)^n]/x
(Geometric sequence with first term (1+x)^n, common ratio (1+x) and there are m terms in the LHS)

Since it is an identity, the coefficient of x^n in LHS should be EQUAL to the coefficient of x^n in RHS.
In LHS, Coefficient of x^n = nCn+(n+1)Cn+......+(n+m)Cn

For RHS, [(1+x)^(n+m+1)-(1+x)^n]/x = (1+x)^(n+m+1)/x - (1+x)^n/x
For the term (1+x)^(n+m+1)/x
= ((n+m+1)C0 + (n+m+1)C1x + (n+m+1)C2x^2+...+(n+m+1)C(n+m+1)x^(n+m+1))/x
=(n+m+1)C0 /x +(n+m+1)C1 +(n+m+1)C2x + (n+m+1)C(n+m+1)x^(n+m)
Hence the coefficient of x^n in (1+x)^(n+m+1)/x is (n+m+1)C(n+1) ...(*)
For the term (1+x)^n/x
=(nC0 + nC1x + nC2x^2 + ... + nCnx^n) / x
=nC0/x + nC1 + nC2x + ... + nCn x^(n-1)
There are no "x^n" term
So the coefficient of x^n in (1+x)^n/x is 0 ...(#)
Combining (*) and (#), the coefficient of x^n in RHS is (n+m+1)C(n+1)

Therefore, nCn+(n+1)Cn+......+(n+m)Cn=(n+m+1)C(n+1)