Question

find the sume of the geometric series by using a formula
1-2+4-8+16-032+64

Answer 1

sumation (-2)^k do you know the formula for geometric series?

Answer 2

no

Answer 3

ok recall the formula

Answer 4

\[\sum_{i=1}^{n}a \left( \frac{1-r^n}{1-r} \right)\]

Answer 5

you have seven terms, and your first term =1

Answer 6

by the way there is a simple way to do this by inducing the terms you provided let me show you

Answer 7

I did the problem and here what i did and can you please tell me if I did it right please and thank you.
1-2+4-8+16-32+64=
-1+4-8+16-32+64=
3-8+16-32+64=
-5+16-32+64=
11-32+64=
-21+64=43
Did i do it right??

Answer 8

he needs to use a GS formula to solve that

Answer 9

ok let me show you a simple way to solve this

Answer 10

\[S_n = 1-2+4-8+16-32+64+...+(-2^n)\]

Answer 11

so a =1 , r=-2, say n=3 the expected value for the sum would be Sn=3, let's check that

Answer 12

recall \[\sum_{i=1}^{n}a \left( \frac{1-r^n}{1-r} \right)\]

Answer 13

\[\sum_{i=1}^{3}(1) \left( \frac{1-(-2)^3}{1-(-2)} \right)=\sum_{i=1}^{3}\left( \frac{9}{3} \right)=3\]

Answer 14

so the sum of the first 3 terms is three let's check it, we have 1-2+4=3, so our answer is correct

Answer 15

now counting the number of terms you provided in the question, which is 7 terms that means now n=7

Answer 16

\[\sum_{i=1}^{7}(1) \left( \frac{1-(-2)^7}{1-(-2)} \right)=\sum_{i=1}^{7}\left( \frac{129}{3} \right)=43\]

Answer 17

let us check 1-2+4-8+16-32+64= 43

Answer 18

This is your final solution, i hope it helped...