How do you determine the number of nth term in this geometric series $sum_{i=1}^n -4^(i-1)=-341$ ?

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Answer 1 · 2017-07-09T02:22:27

$sum_{i=1}^n -4^(i-1)=-341$

$=>sum_{i=1}^n 4^(i-1)=341$

$=>(4^0+4^1+4^2+--------4^(n-1)=341)$

$=>(4^0(4^n-1))/(4-1)=341$

$=>4^n=341xx3+1=1024=4^5$

$=>n=5$

Answer 2 · 2017-07-09T02:27:47

$n=5$

$sum_"i=1"^n -4^(i-1)$ is a geometric sum.

The first term $(a_1) = -4^0 =-1$

The common ratio $(r) = (-4^1)/(-1) = 4$

The sum of the first n terms $(S_n)$ is given by:

$S_n = (a_1(1-r^n))/(1-r)$

In this question we are told that $S_n = -341$ and asked to solve for $n$ .

$-341 = (-1(1-4^n))/(1-4)$

$(1-4^n)/3 = 341$

$1-4^n = -1023$

$4^n = 1024$

$2^(2n) = 2^10$

$2n =10 -> n=5$

How do you determine the number of nth term in this geometric series sum_{i=1}^n -4^(i-1)=-341sum_{i=1}^n -4^(i-1)=-341?