Question

find the numerical answer to the summation given below.\\(\sum_{n=2}^{62} (4n + 7)\\)

Explanation:

Step1: Recall the sum of arithmetic series formula

The sum of an arithmetic series \(\sum_{k = 1}^{n} a_k\) is given by \(S_n=\frac{n(a_1 + a_n)}{2}\), where \(a_1\) is the first term and \(a_n\) is the \(n\)th term. First, we can rewrite \(\sum_{n = 2}^{62}(4n + 7)\) as \(\sum_{n = 1}^{62}(4n + 7)-\sum_{n = 1}^{1}(4n + 7)\).

Step2: Find the general term of the arithmetic sequence

The general term of the sequence \(a_n=4n + 7\). For an arithmetic sequence, the common difference \(d\) is the coefficient of \(n\), so \(d = 4\).

Step3: Calculate \(\sum_{n = 1}^{62}(4n + 7)\)

First, find \(a_1\) when \(n = 1\): \(a_1=4\times1 + 7=11\).
Find \(a_{62}\) when \(n = 62\): \(a_{62}=4\times62+7=248 + 7=255\).
The number of terms \(n = 62\). Using the sum formula \(S=\frac{n(a_1 + a_n)}{2}\), we have \(S_{62}=\frac{62\times(11 + 255)}{2}=\frac{62\times266}{2}=62\times133 = 8246\).

Step4: Calculate \(\sum_{n = 1}^{1}(4n + 7)\)

When \(n = 1\), \(4n+7=11\), so \(\sum_{n = 1}^{1}(4n + 7)=11\).

Step5: Find \(\sum_{n = 2}^{62}(4n + 7)\)

Subtract the two sums: \(8246-11 = 8235\).

Answer:

8235