Summation of a finite sequence

Question

This question is linked from my previous question: Summation of a sequence?

Given the sequence:

$$ a_n = 0.9^{n-1}a_1(1+d+d^2+...d^{n-1}) $$ and $a_1=100$ , $d= 1.5$

How to form an equation to find: $$ \sum^5_{n=1}a_n $$ Please help me.

Answer 1

What you have written is $$a_n=90\sum^r_{n=1}d^{n-1}$$.

Now taking summation on both sides

$$\sum^5_{r=1}a_n=90\sum^5_{r=1}\sum^r_{n=1}d^{n-1} $$

Hence it is equal to

$\sum^5_{r=1}a_n=90*(d^0+d^0+d^1+d^0+d^1+d^2+d^0+d^1+d^2+d^3+d^0+d^1+d^2+d^3+d^4) $

$\sum^5_{r=1}a_n=90*(5d^0+4d^1+3d^2+2d^3+d^4) $

Now you may substitute values.

Answer 2

You just have to rewrite some terms, it's not very difficult. Just remember the geometric sum. We first replace $$\tag{1} 1+d+d^2 +d^3 + \ldots + d^{n-1} = \frac{d^n -1 }{d-1} $$ so now the sequence is $$ a_n = 0.9^{n-1} \cdot a_1 \cdot \frac{d^n -1 }{d-1} $$ Let's replace $0.9$ by $q$ and then make it $q^{n-1} = q^n/q$. Then we have $$ a_n = \frac{a_1}{q (d-1)} \cdot q^n (d^n -1) $$ the part $a_1/(q(d-1))$ is a constant, so we can just name it as $C$ and then we have $$ a_n = C (qd)^n - C q^n $$ When we take the sum over this, we can use Equation (1) again: $$ \begin{split} \sum_{n=0}^N a_n =& C \sum_{n=1}^N (qd)^N - C\sum_{n=1}^n q^n\\ =& C \frac{(qd)^{N+1}-1}{qd -1} - C\frac{q^{N+1}-1}{q-1} \end{split} $$ We can now put the value of $C$ back into its place to get the final result, $$ \sum_{n=0}^N a_n = \frac{a_1}{q (d-1)} \frac{(qd)^{N+1}-1}{qd -1} - \frac{a_1}{q (d-1)}\frac{q^{N+1}-1}{q-1} $$ In your case, you start from $1$ instead of zero, and $N=5$ so you'll have to adjust the equation accordingly.

Answer 3

So , $$a_n=0.9^{n-1}a_1\frac{(d^n-1)}{d-1}$$ We need to find : $$\sum_{n=1}^{5}a_n=\sum_{n=1}^{5}\left(0.9^{n-1}a_1\frac{(d^n-1)}{d-1}\right)=200\sum_{n=1}^{5}0.9^{n-1}\times(1.5^n-1)$$ $$\Rightarrow200\left(\sum_{n=1}^{5}\frac{1.35^n}{0.9}-\sum_{n=1}^{5}0.9^{n-1}\right)=200\left(\frac{10}{9}\times\frac{1.35(1.35^5-1)}{0.35}+\frac{(0.9^5-1)}{0.1}\right)$$

From here you can find , the answer .