Problem 42 - College Algebra

Question

Find each indicated sum. $$\sum_{i=1}^{5} \frac{(i+2) !}{i !}$$

Step-by-Step Solution

Answer

The sum of the series is 110.

Step 1: Factorize the terms in the sum

Factorize the expression under the sum. In this case, \(\frac{(i+2) !}{i !}\) simplifies to \((i+2)(i+1)\) since \((i+2)!\) is equivalent to \((i+2)(i+1)(i)!\) and \(i!\) cancels out.

Step 2: Calculate the sum for each i from 1 to 5

Now calculate the value of \((i+2)(i+1)\) for each value of \(i\) from 1 to 5 and add them up. So calculation is as follows: when \(i=1\), the result = \((1+2)(1+1) = 6\), when \(i=2\), the result = \((2+2)(2+1) = 12\), when \(i=3\), the result = \((3+2)(3+1) = 20\), when \(i=4\), the result = \((4+2)(4+1) = 30\), and when \(i=5\), the result = \((5+2)(5+1) = 42\). Now, add these all up i.e., \(6 + 12 + 20 + 30 + 42 = 110\)

Step 3: Formulate the answer

After the calculation, the final answer becomes apparent, which is the total sum of the series.