Find the 7th partial sum of summation of 6 open parentheses 3 close parentheses to the I minus 1 power from 1 to infinity

Question

asked Sep 11, 2017 114k views

2 Answers

Rakesh B E · Answer 1 · 2017-09-15T14:16:15+0000

Answer: 6558

Explanation:

Given expression,

$\sum_(i=1)^(\infty) 6(3)^(i-1)$

Since, for the 7th partial sum of this summation.

we take i from 1 to 7.

Thus, the required sum is,

$\sum_(i=1)^(7) 6(3)^(i-1)=6(3)^0+6(3)^1+6(3)^2+6(3)^3+6(3)^4+6(3)^5+6(3)^6$

$\sum_(i=1)^(7) 6(3)^(i-1)=6+18+54+162+486+1458+4374$

$\sum_(i=1)^(7) 6(3)^(i-1)=6558$