Question

Fill in the blanks to complete each sum formula:

$\sum _{k=1}^{n}k=\underline{}$

Short answer

The complete sum formula is$\sum _{k=1}^{n}k=\underline{\frac{n(n+1)}{2}}.$

Step-by-step solution

Step 1. Given information.

Given incomplete sum formula is following.

$\sum _{k=1}^{n}k=\underline{}$

Step 2. proof

Write the expanded form of sigma notation.

$\sum _{k=1}^{n}k=1+2+3+\cdots (n-1)+n$

The sum of the first and the last term is $1+n.$

Similarly, the sum of the second and second last terms will be $2+(n-1)=n+1$

The number of pairs whose sum is $1+n$will be$\frac{n}{2}.$

So the sum of all pairs will be$\frac{n}{2}(n+1)=\frac{n(n+1)}{2}$

the sum of all pairs is equal to sigma notation.

so$\sum _{k=1}^{n}k=\frac{n(n+1)}{2}$