Find the sum of the summation of 5 is the upper limit, I=1 and 3(-4)^i
To find the sum of the summation with an upper limit of 5, i=1, and the term 3(-4)^i:
S = Σ 3(-4)^i where i=1 to 5
This can be written as:
S = 3(-4)^1 + 3(-4)^2 + 3(-4)^3 + 3(-4)^4 + 3(-4)^5
Now calculate each term:
- First term: 3(-4)^1 = 3(-4) = -12
- Second term: 3(-4)^2 = 3(16) = 48
- Third term: 3(-4)^3 = 3(-64) = -192
- Fourth term: 3(-4)^4 = 3(256) = 768
- Fifth term: 3(-4)^5 = 3(-1024) = -3072
Now sum up all the terms:
S = -12 + 48 - 192 + 768 - 3072 = -2460
Therefore, the sum of the given series with an upper limit of 5 is -2460.