How do you find the 4-th partial sum of the infinite series $sum_(n=1)^oo(3/2)^n$ ?

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Answer 1 · 2018-05-13T12:56:30

$195/16$

When dealing with a sum, you have a sequence that generates the terms. In this case, you have the sequence

$a_n = (3/2)^n$

Which means that $n$ -th term is generates by raising $3/2$ to the $n$ -th power.

Moreover, the $n$ -th partial sum means to sum the first $n$ terms from the sequence.

So, in your case, you're looking for $a_1+a_2+a_3+a_4$ , which means

$3/2 + (3/2)^2 + (3/2)^3 + (3/2)^4$

You may compute each term, but there is a useful formula:

$sum_{i=1}^n k^i= \frac{k^{n+1}-1}{k-1}$

So, in your case

$sum_{i=0}^4 (3/2)^i= \frac{(3/2)^{5}-1}{3/2-1} = 211/16$

Except you are not including $a_0 = (3/2)^0 = 1$ in your sum, so we must subtract it:

$sum_{i=0}^4 (3/2)^i = sum_{i=1}^4 (3/2)^i - 1 = 211/16 - 1 = 195/16$

Answer 2 · 2018-05-13T12:57:15

$sum_1^(n=4)(3/2)^n = 12.1875$

This is geometric progression series of which

first term is $a=3/2=1.5$ , common ratio is $r=1.5$

$4$ th partial sum ; i.e $n=4$

Sum $S= a * (r^n-1)/(r-1)= 1.5 * ((1.5^4-1)/(1.5-1))=12.1875$

$sum_1^(n=4)(3/2)^n = 12.1875$ [Ans]

How do you find the 4-th partial sum of the infinite series sum_(n=1)^oo(3/2)^nsum_(n=1)^oo(3/2)^n ?