Question

Evaluate the summation of 25 times 0.3 to the n plus 1 power, from n equals 2 to 10..

Answer 1

We are asked to evaluate the summation of 25 times 0.3 to the n plus 1 power, from n equals 2 to 10. In this case, we use a calculator with summation powers so as to accurately get the answer. Using a calculator, the asnwer is equal to 0.9643. 

Answer 2

Answer:

$\sum_{2}^{10}25(0.3)^{n+1}=0.9642375$

Step-by-step explanation:

We have to evaluate the expression:

$\sum_{2}^{10}25(0.3)^{n+1}$

i.e. it could also be written as:

$25\sum_{2}^{10}(0.3)^{n+1}$

i.e. we need to evaluate:

$25[(0.3)^3+(0.3)^4+(0.3)^5+(0.3)^6+(0.3)^7+(0.3)^8+(0.3)^9+(0.3)^{10}+(0.3)^{11}]$

Hence, this could be written as:

$=25\times (0.3)^3[1+0.3^1+0.3^2+0.3^3+0.3^4+0.3^5+0.3^6+0.3^7+0.3^8]$

Now, the series inside the parenthesis is a geometric series with first term as 1 and common ration as 0.3.

Hence, we could apply the summation of finite geometric series and get the answer.

We know that the sum of geometric series with n terms and common ratio less than 1  is calculated as:

$S_n=a\times (\dfrac{1-r^n}{1-r})$

Here a=1 and r=0.3

Hence the sum of geometric series is:

$S_9=1\times (\dfrac{1-0.3^9}{1-0.3})\\\\S_9=1.4285$

Hence, the final evaluation is:

$=25\times (0.3)^3\times 1.4285\\\\=0.9642375$

Hence,

$\sum_{2}^{10}25(0.3)^{n+1}=0.9642375$