Coefficient in binomial expansion

Question

Whats is the coefficient in front of $x^{50}$ in the binomial expansion $$(1+x)^{1000} + (1 + x)^{999}\cdot x^1 + ... + (1+ x)^1\cdot x^{999} + x^{1000}$$

Answer 1

Hint:

$$\sum_{r=0}^{1000}(1+x)^{1000-r}x^r$$ is a finite geometric series

$$=(1+x)^{1000}\cdot\dfrac{1-\left(\dfrac x{x+1}\right)^{1000+1}}{1-\dfrac x{x+1}}$$

Answer 2

From lab bhattacharjee's answer, we can simplify further:

$$(1+x)^{1000} \cdot \frac{1-\big(\frac{x}{x+1}\big)^{1001}}{\frac{1}{x+1}}$$ $$=(1+x)^{1001} - (1+x)^{1001} \cdot \big(\frac{x}{x+1}\big)^{1001}$$ $$=(1+x)^{1001} - x^{1001}$$

Can you then use the binomial expansion of $(1+x)^{1001}$ to find the coefficient of $x^{50}$?

Answer 3

The required coefficient is $\binom {1000}{50}+\binom {999}{49}+\cdots \binom {950}{0}$

Since $\binom {n}{r} = \binom {n}{n-r}$, the above sum can be written as

$\binom {1000}{950}+\binom {999}{950}+\cdots \binom {950}{950}$ which by the Hockey Stick Identity is $\binom {1001}{951}$