Question

Find the 6th partial sum of summation of 7 open parentheses 4 close parentheses to the I minus 1 power from 1 to infinity

Answer 1

Answer: option C. 9,555.

Justification:

1) The summation is:

6
∑  7(4)^ (i - 1)
i=1

2) Substitute with the values of i from 1 to 6 (note that 7 is a common factor of all the terms) =>

7 { 4^(1-1) + 4^(2-1) + 4^(3-1) + 4^(4-1) + 4^(5-1) + 4^(6-1) } =

= 7 { 4^0 + 4^1 + 4^2 + 4^3 + 4^4 + 4^5 } = 7 { 1 + 4 + 16 + 64 + 256 + 1024} =

7{1365} = 9,555.

Answer: 9,555