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

\[\sum_{i=1}^{\infty} 6(3)^i-1\]

Answer 2

it's suppose to be 6(3)^i-1

Answer 3

2,184
6,558
27,993
167,961

Answer 4

I got B :)

Answer 5

Me too

Sn = a*(1-r^n)/(1-r)
S7 = 6*(1-3^7)/(1-3)
S7 = 6,558

Answer 6

Yayyy! Twinning hahah :D Thanks

Answer 7

no problem