Powers of 1/2

Form $S_{n}=\frac{a_{1}(1-r^{n})}{1-r}$ when r is not equal to 1 in this case $r =\frac{1}{2}$ then $S_{n}=\frac{\frac{1}{2}(1-(\frac{1}{2})^{7})}{1-\frac{1}{2}} = \frac{127}{128}$ 127 is a prime number then $\frac{127}{128}$ is co-prime numbers then a+b = 127+128=255 Ans = 255

Its pretty simple. The denominators, if you simplify them becomes 2,4,8,16,32,64 and 128 respectively. Now all you need to do is simple addition of multiple fractions which gives the answer as 127/128. Now, add 127 and 128 which gives you your final answer, that is 255.

We use here the formula for sum of geometric progression(GP), S $=a.\frac{r^{n}-1}{r-1}$ where r=Common ratio and n=No. of terms. In this problem, n=7 and r=1/2.

S $=\frac{1}{2}+\frac{1}{2^{2}}+.....+\frac{1}{2^{7}}$

S $=\frac{1}{2}\times \frac{(\frac{1}{2})^{7}-1}{\frac{1}{2}-1}$

S $=\frac{1}{2}\times \frac{\frac{1}{2^{7}}-1}{\frac{-1}{2}}$

S $=1-\frac{1}{2^{7}} = \frac{2^{7}-1}{2^{7}} = \frac{127}{128}$

Now, S is in $\frac{a}{b}$ form with a and b coprime, a=127 and b=128. So, $a+b=127+128=\boxed{255}$

1/2+1/4+1/8+1/16+1/32+1/64+1/128... so a=1 b=254. i.e 1+254=255

We can take a common factor $\frac{1}{2}$ out to make the solution easier!

$S = \frac{1}{2^{1}} + \frac{1}{2^{2}} + \frac{1}{2^{3}} + ... + \frac{1}{2^{7}}$

$S = \frac{1}{2^{1}} \times (1 + \frac{1}{2^{1}} + \frac{1}{2^{2}} + \frac{1}{2^{3}} + ... + \frac{1}{2^{6}})$

$S = \frac{1}{2} \times (1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \frac{1}{32} + \frac{1}{64})$

$S = \frac{1}{2} \times (\frac{64+32+16+8+4+2+1}{64})$

$S = \frac{1}{2} \times (\frac{127}{64})$

$S = \frac{127}{128}$

So, $a = 127, b = 128$

Therefore, the answer: $a + b = 127 + 128 = \boxed{255}$

s/2-s=-s/2

-s/2=1/2^8-1/2=-127/256

s=-127/256(-2)=127/128

127+128=255

(1/2)+(1/2^2)+(1/2^3)+(1/2^4)+(1/2^5)+(1/2^6)+(1/2^7)=255