Question

find the indicated sum.
\\(\sum_{i=3}^{7} \frac{i!}{(i - 2)!}\\)
\\(\sum_{i=3}^{7} \frac{i!}{(i - 2)!} = \square\\)

Explanation:

Step1: Understand the summation

We need to calculate the sum \(\sum_{i = 3}^{7}\frac{i!}{(i - 2)!}\). This means we will substitute \(i = 3\), \(i = 4\), \(i = 5\), \(i = 6\), and \(i = 7\) into the expression \(\frac{i!}{(i - 2)!}\) and then add up the results.

Step2: Simplify the general term

Recall that \(n! = n\times(n - 1)\times(n - 2)\times\cdots\times1\). So, \(\frac{i!}{(i - 2)!}=\frac{i\times(i - 1)\times(i - 2)!}{(i - 2)!}\). The \((i - 2)!\) terms cancel out, leaving us with \(i\times(i - 1)=i^{2}-i\).

Step3: Calculate for each \(i\)

• For \(i = 3\): \(3^{2}-3 = 9 - 3=6\)
• For \(i = 4\): \(4^{2}-4 = 16 - 4 = 12\)
• For \(i = 5\): \(5^{2}-5 = 25 - 5 = 20\)
• For \(i = 6\): \(6^{2}-6 = 36 - 6 = 30\)
• For \(i = 7\): \(7^{2}-7 = 49 - 7 = 42\)

Step4: Sum the results

Now, we add up these values: \(6 + 12+20 + 30+42\).
First, \(6+12 = 18\). Then, \(18 + 20 = 38\). Next, \(38+30 = 68\). Finally, \(68 + 42 = 110\).

Answer:

\(110\)