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,
\(\sum_{i=1}^{7} 6(3)^{i-1}=6(3)^0+6(3)^1+6(3)^2+6(3)^3+6(3)^4+6(3)^5+6(3)^6\)
\(\sum_{i=1}^{7} 6(3)^{i-1}=6+18+54+162+486+1458+4374\)
\(\sum_{i=1}^{7} 6(3)^{i-1}=6558\)
