Question

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

Answer 1

Answer: 6558

Step-by-step 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,

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

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

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