Question

i need help please what is the sum of the finite geometric series??
1)1+2+4+...+128
2)3+6+12+24+48+...+768


thanks :)

Answer 1

anybody ?? please :(

Answer 2

The sum of a finite geometric series is given by
\[\sum_{i=1}^{n} a_i = a_1 \frac{1-r^n}{1-r}\]
Where
\[a_1 = \text{The first term of your geometric series}\]
\[r = \text{The common ratio of your geometric series}\]
\[n = \text{The number of terms \in your geometric series}\]
So for the first geometric series, looking at each term:
\[1 + 2 + 4 + \ldots\]
What do we multiply 1 by to get 2? What do we multiply 2 by to get 4? Notice that it is the same number--that is our common ratio, r.

Answer 3

thank you very much!! (;

Answer 4

i did not get it but thanks anyway

Answer 5

Which part is confusing you? I'll try to explain it better.

Answer 6

1)

\[S=1+2+4+\ldots+64+128\]
\[2S=2(1+2+4+\ldots+64+128)=(2+4+8+\ldots+128+256)\]
\[2S-S=S=(2+4+8+\ldots+128+256)-(1+2+4+\ldots+64+128)\]
\[S=(\cancel{2}+\cancel{4}+\cancel{8}+\ldots+\cancel{128}+256)-(1+\cancel{2}+\cancel{4}+\ldots+\cancel{64}+\cancel{128})\]
\[S=256-1=255\]

Now try with 2)