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

User Edorian
by
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2 Answers

4 votes

Answer:b

Step-by-step explanation:test

User Eustace
by
8.2k points
1 vote

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

User Rakesh B E
by
8.1k points
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