123456
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Geometric Series

So, what if we want to find this sum?

the summation of ( 4( 1 / 2 )^( k - 1 ) ) as k goes from 1 to 10

Luckily, there's a formula!

To find the sum of the first n terms of a geometric sequence:

Sn = the summation of ( a1 * r^( k -1 ) ) as k goes from 1 to n = a1( ( 1 - r^n ) / ( 1 - r ) )

Let's use it:

S10 = the summation of ( 4 * ( 1 / 2 )^( k - 1 ) ) = 4( ( 1 - ( 1 / 2 )^10 ) / ( 1 - ( 1 / 2 ) ) = 8( 1 - ( 1 / 2 )^10 ) = ( 1023 / 128 ) ... n is 10 , a1 is 4 , r is ( 1 / 2 )

Check to make sure the formula works by adding these up:

4 + 2 + 1 + ( 1 / 2 ) + ( 1 / 4 ) + ( 1 / 8 ) + ( 1 / 16 ) + ( 1 / 32 ) + ( 1 / 64 ) + ( 1 / 128 ) = ___

 

TRY IT:

Find the sum of the first 15 terms of

5 , -10 , 20 , -40 , 80 , ...