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

Asked by maham237 @ in Mathematics viewed by 418 People

Answered by maham237 @


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


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