The coefficient of $$x^{48}$$ in $$ (1+x) + 2(1+x)^{2}+3(1+x)^{3}+....+100(1+x)^{100} $$ is equal to

To find the coefficient of $$x^{48}$$ in $$S = \sum_{k=1}^{100} k(1+x)^k$$, we use the derivative of a geometric series.

We know that for a geometric series $$\sum_{k=0}^{n} y^k = \frac{y^{n+1}-1}{y-1}$$. Differentiating both sides with respect to $$y$$:

$$\sum_{k=1}^{n} ky^{k-1} = \frac{d}{dy} \left[ \frac{y^{n+1}-1}{y-1} \right]$$

Multiplying by $$y$$ to match our series form:

$$\sum_{k=1}^{n} ky^k = y \frac{(n+1)y^n(y-1) - (y^{n+1}-1)}{(y-1)^2}$$

Let $$y = (1+x)$$ and $$n = 100$$. Since $$y-1 = x$$:

$$S = \frac{(1+x) [101(1+x)^{100}(x) - (1+x)^{101} + 1]}{x^2}$$

$$S = \frac{101(1+x)^{101}}{x} - \frac{(1+x)^{102}}{x^2} + \frac{1+x}{x^2}$$

• Term 1: Coefficient of $$x^{48}$$ in $$\frac{101(1+x)^{101}}{x}$$ is $$101 \binom{101}{49}$$.
• Term 2: Coefficient of $$x^{48}$$ in $$\frac{(1+x)^{102}}{x^2}$$ is $$\binom{102}{50}$$.
• Term 3: $$\frac{1+x}{x^2}$$ has no $$x^{48}$$ term.

Total coefficient $$= 101 \binom{101}{49} - \binom{102}{50}$$.

Using the identity $$\binom{n}{r} = \binom{n-1}{r} + \binom{n-1}{r-1}$$ on the second term:

$$= 101 \binom{101}{49} - \left[ \binom{101}{50} + \binom{101}{49} \right]$$

$$= 100 \binom{101}{49} - \binom{101}{50}$$