General Formula Derivation

Question

Please derive this

$1+2+4+........+\frac{n}{4}+\frac{n}{2}=2^{log n}-1$

Comments

$2^{0} + 2^{1} + .....................2^{(k-1)}$

using gp : $\frac{2^{(m+1)}-1}{(2-1)}$ = $\frac{2^{(k-1+1)} -1 }{1}$ = $2^{k}-1$

and$ \frac{n}{2^k}$ =1 => $n=2^{k}$

=>$k =logn$. (taking log both side)

therefore : $2^{logn} -1$..

---

I'm not getting please explain?

---

now check

Answer 1

First term a = 1 and ratio r = 2