Q. 14 - Calculus

Question

Split the sum $\sum _{k=4}^{11} \sqrt{k}$ into three sums, each in sigma notation, where the first sum has two terms and the last two sums each have three terms. 

Step-by-Step Solution

Answer

The solution is:$\sum _{k-4}^{11}\sqrt{k}=\sum _{k-4}^5\sqrt{k}+\sum _{k-6}^8\sqrt{k}+\sum _{k-9}^{11}\sqrt{k}$

Step 1. Given Information.

Consider the given question,

$\sum _{k=4}^{11} \sqrt{k}$

Step 2. Calculation

Consider the given sum-

$\sum _{k-4}^{11}\sqrt{k}$

Here k varies from $4 to 11$

So, it can be divided into three sums as shown:

$\sum _{k-4}^{11}\sqrt{k}=\sum _{k-4}^5\sqrt{k}+\sum _{k-6}^8\sqrt{k}+\sum _{k-9}^{11}\sqrt{k}$

Now, $\sum _{k-4}^5$ has two terms in the sums as $\sqrt{4}+\sqrt{5}$

$\sum _{k-6}^8\sqrt{k}$ has three terms in the sum as $\sqrt{6}+\sqrt{7}+\sqrt{8}$

$\sum _{k-9}^{11}\sqrt{k}$ has three terms in the sum as $\sqrt{9}+\sqrt{10}+\sqrt{11}.$