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

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Answer 1 · 2017-07-09T02:22:27

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

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

$\Rightarrow (4^{0} + 4^{1} + 4^{2} \pm - - - - - - - 4^{n - 1} = 341)$ $=>(4^0+4^1+4^2+--------4^(n-1)=341)$

$\Rightarrow \frac{4^{0} (4^{n} - 1)}{4 - 1} = 341$ $=>(4^0(4^n-1))/(4-1)=341$

$\Rightarrow 4^{n} = 341 \times 3 + 1 = 1024 = 4^{5}$ $=>4^n=341xx3+1=1024=4^5$

$\Rightarrow n = 5$ $=>n=5$

Answer 2 · 2017-07-09T02:27:47

$n = 5$ $n=5$

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

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

The common ratio $(r) = \frac{- 4^{1}}{- 1} = 4$ $(r) = (-4^1)/(-1) = 4$

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

$S_{n} = \frac{a_{1} (1 - r^{n})}{1 - r}$ $S_n = (a_1(1-r^n))/(1-r)$

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

$- 341 = \frac{- 1 (1 - 4^{n})}{1 - 4}$ $-341 = (-1(1-4^n))/(1-4)$

$\frac{1 - 4^{n}}{3} = 341$ $(1-4^n)/3 = 341$

$1 - 4^{n} = - 1023$ $1-4^n = -1023$

$4^{n} = 1024$ $4^n = 1024$

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

$2 n = 10 \to n = 5$ $2n =10 -> n=5$

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