Find $a_1$ given that $(1+x)^{100} = \sum_{i=0}^{100} a_ix^i$

Question

If $(1+x)^{100} = \sum_{i=0}^{100} a_ix^i$, then $a_1$ is ..

The options are $1$, $2$, $99$ or $100$. I'm sure the problem is trivial, but I just don't understand what is meant.

Answer 1

The question is asking you to expand the binomial series for $(1+x)^{100}$. The right-hand side is sigma-notation, which when expanded gives you $\sum_{i=0}^{100}{a_i x^i} = a_0 + a_1 x + a_2 x^2 + \ldots + a_{100} x^{100}$. Since $a_1$ is the coefficient of $x^1 = x$, what is that coefficient?

Answer 2

By binomial theorem $a_i=\binom{n}{i}$ then: $$a_1=\binom{100}{1}=\frac{100!}{(100-1)!\cdot1!}=100$$

Answer 3

Alternatively, try a Taylor Series. If $f(x)=(1+x)^{100}$, then the coefficient of $x$ is equal to $\frac{f'(0)}{1!}=f'(0)$. $f'(x)=100(1+x)^{99}$, so $f'(0)=100=a_1$.