Answer
The sum of the first three terms is $2.495$.
Work Step by Step
Rewrite $1.01$ as $1+0.01$, then expand the left side of the inequality using the Binomial Theorem:
$$\begin{align*}
(1.01)^{100}&=(1+0.01)^{100}\\
&=\binom{100}{0}1^{100}+\binom{100}{1}1^{99}(0.01)+\binom{100}{2}1^{99}(0.01)^2+\dots+0.01^{100}\\
&=1+100(0.01)+4950(0.0001)+\dots+0.01^{100}\\
&=1+1+0.495+\dots+0.01^{100}\\
&=2.495+\dots+0.01^{100}.
\end{align*}$$
Since the sum of the first three terms of the expansion is greater than $2$ and the rest of the terms are positive, we have:
$$1.01^{100}=2.495+\dots+0.01^{100}>2.$$
