Sequence; sequence; logarithm!

As given, we have $a_{n+1}=2^{n}\cdot a_{n}$ , so: $\log_{2}(a_{100})=\log_{2}(2^{99}\cdot a_{99})=\log_{2}(2^{99}\cdot 2^{98}\cdot a_{98})=\cdots$ $=\log_{2}(\displaystyle \prod_{k=1}^{99} 2^{k} \cdot a_{1})=\log_{2}(2^{\sum_{k=1}^{99} k})$ $=\displaystyle \sum_{k=1}^{99} k \cdot \log_{2}(2)=\frac{99\cdot 100}{2}=(90+9)\cdot 50=4500+450=4950$

We know that: $\frac{a_{n+1}}{a_{n}}=2^{n}$ , $\frac{a_{n}}{a_{(n-1}}=2^{n-1}$ ,..., $\frac{a_{2}}{a_{1}}=2^{1}$ We multiply all above in order to reach: $\frac {a_{n+1}} {a_{n}}$ $* \frac{a_{n}}{a_{(n-1)}}$ $* ... * \frac{a_{2}}{a_{1}} =2^{n+(n-1)+...+1}$ . $\frac {a_{n+1}} {a_{n}} = 2^{ \frac{n(n+1)}{2}}$ , since $1+2+...+n=\frac {n(n+1)}{2}$ . Then $a_{n+1}=2^{\frac{n(n+1)} {2} }$ . $a_{1}=2^{\frac{(n+1)n}{2}}$ Substituting $n+1$ for $n$ we get $a_{n}=2^{\frac{n(n-1)}{2}}$ and $log_{2} (a_{n})= \frac {(n-1)n}{2}$ . We need $log_{2} (a_{100})= \frac{100.(100-1)}{2}=50.99=4950$

$a_{n+1} = 2^n \times a_n$

$a_1 = 1 = 2^0$

$a_2 = 2^1 \times a_1 = 2^1 \times 2^0 = 2^1 = 2^{0 + 1}$

$a_3 = 2^2 \times a_2 = 2^2 \times 2^1 = 2^3 = 2^{0 + 1 + 2}$

$a_4 = 2^3 \times a_3 = 2^3 \times 2^3 = 2^6 = 2^{0 + 1 + 2 + 3}$

$a_5 = 2^4 \times a_4 = 2^4 \times 2^6 = 2^{10} = 2^{0 + 1 + 2 + 3 + 4}$

$a_6 = 2^5 \times a^5 = 2^5 \times 2^{10} = 2^{15} = 2^{0 + 1 + 2 + 3 + 4 + 5}$

We can generalize this $:$

$a_n = 2^{1 + 2 + 3 + ... + (n - 1)}$

Thus,

$a_{100} = 2^{1 + 2 + 3 + ... + 99}$

Recall that the sum of the first $n$ numbers $\displaystyle \sum_{k=0}^n {k} = \frac 12 (n)(n + 1).$

$1 + 2 + 3 + ... + 99 = \frac {100 \times 99}{2} = 4950$

$a_{100} = 2^{1 + 2 + 3 + ... + 99} = 2^{4950}$

$\log_2 (a_{100}) = \log_2 {2^{4950}} = 4950$

$\frac{a_{n+1}}{a_n} = 2^{n}$ ; and $a_1 = 1$ .

$n =1 \Rightarrow \frac{a_2}{a_1} = 2 \rightarrow a_2 = 2^{1}$

$n = 2 \Rightarrow \frac{a_3}{a_2} = 2^{2} \rightarrow a_3 = 2^{3}$

$n = 3 \Rightarrow \frac{a_4}{a_3}= 2^{3} \rightarrow a_4 = 2^{6}$

$n= 4 \Rightarrow \frac{a_5}{a_4} = 2^{4} \rightarrow a_5 = 2^{10}$

We now can see that $a_n = 2^{\frac{n(n-1)}{2}}$ .

$a_{100} = 2^{4950}$ .

Hence $log_2 (a_{100}) = 4950$