Answer
please see step-by-step
Work Step by Step
Rewriting $(1.01 )^{100} =(1+0.01)^{100}$,
which , by the Binomial Theorem, expands into 101 terms,
all of form $ \left(\begin{array}{l}
n\\
r
\end{array}\right)a^{r}b^{n-r}$,
all positive,
where the first three are
$ \left(\begin{array}{l}
100\\
0
\end{array}\right)1^{100}\cdot(0.01)^{0}=1\cdot 1\cdot 1=1$
$ \left(\begin{array}{l}
100\\
1
\end{array}\right)1^{99}\cdot(0.01)^{1}=100\cdot 1\cdot 0.01=1$
$ \displaystyle \left(\begin{array}{l}
100\\
2
\end{array}\right)1^{98}\cdot(0.01)^{2}=\frac{100\cdot 99}{2}\cdot 1\cdot 0.0001=0.495$
The sum of the first three terms in the expansion of $(1+0.01)^{100}$ is greater than 2,
and since none of the remaining 98 terms is negative, it follows that the whole sum is greater than 2.
$(1+0.01)^{100}= (1.01 )^{100} > 2$
