Question

find the indicated sum.
\\(\displaystyle \sum_{i=1}^{3} i(i + 5)\\)
\\(\displaystyle \sum_{i=1}^{3} i(i + 5) = \square\\) (simplify your answer.)

Explanation:

Step1: Expand the term

First, expand \(i(i + 5)\) using the distributive property (also known as the distributive law of multiplication over addition). The formula for the distributive property is \(a(b + c)=ab+ac\). Applying this to \(i(i + 5)\), we get:

\(i(i + 5)=i\times i+i\times5=i^{2}+5i\)

Step2: Substitute \(i = 1,2,3\) and sum

Now, we need to find the sum \(\sum_{i = 1}^{3}i(i + 5)=\sum_{i = 1}^{3}(i^{2}+5i)\). This is equivalent to finding the sum when \(i = 1\), \(i = 2\), and \(i = 3\) and then adding those results together.

• When \(i = 1\):

Substitute \(i = 1\) into \(i^{2}+5i\):
\(1^{2}+5\times1=1 + 5=6\)

• When \(i = 2\):

Substitute \(i = 2\) into \(i^{2}+5i\):
\(2^{2}+5\times2=4 + 10=14\)

• When \(i = 3\):

Substitute \(i = 3\) into \(i^{2}+5i\):
\(3^{2}+5\times3=9 + 15=24\)

Step3: Sum the results

Now, sum the three results we got: \(6+14 + 24\).

First, add \(6\) and \(14\): \(6+14 = 20\).

Then, add \(20\) and \(24\): \(20+24=44\).

Alternatively, we can also use the properties of summation \(\sum_{i = 1}^{n}(a_{i}+b_{i})=\sum_{i = 1}^{n}a_{i}+\sum_{i = 1}^{n}b_{i}\). So \(\sum_{i = 1}^{3}(i^{2}+5i)=\sum_{i = 1}^{3}i^{2}+5\sum_{i = 1}^{3}i\).

We know the formula for the sum of the first \(n\) positive integers: \(\sum_{i = 1}^{n}i=\frac{n(n + 1)}{2}\), and the formula for the sum of the squares of the first \(n\) positive integers: \(\sum_{i = 1}^{n}i^{2}=\frac{n(n + 1)(2n + 1)}{6}\).

For \(n = 3\):

\(\sum_{i = 1}^{3}i^{2}=\frac{3\times(3 + 1)\times(2\times3+ 1)}{6}=\frac{3\times4\times7}{6}=\frac{84}{6}=14\)

\(5\sum_{i = 1}^{3}i=5\times\frac{3\times(3 + 1)}{2}=5\times\frac{12}{2}=5\times6 = 30\)

Then, \(\sum_{i = 1}^{3}i^{2}+5\sum_{i = 1}^{3}i=14 + 30=44\)

Answer:

\(44\)