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.

Short 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-by-step solution

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 $4to11$

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$\sqrt{9}+\sqrt{10}+\sqrt{11}.$