Question

find the numerical answer to the summation given below.
\\(\sum_{n=2}^{68} (3n + 2)\\)

Explanation:

Step1: Recall the sum of arithmetic series formula

The sum of an arithmetic series \(\sum_{k = 1}^{n} (a + (k - 1)d)\) is \(S_n=\frac{n}{2}(2a + (n - 1)d)\) or \(S_n=\frac{n}{2}(a_1 + a_n)\), where \(a_1\) is the first term and \(a_n\) is the \(n\)th term. First, we need to adjust our summation from \(n = 2\) to \(n = 68\) to be in terms of a standard arithmetic series. Let's find the first term when \(n = 2\): \(a_1=3(2)+2 = 8\). The last term when \(n = 68\): \(a_{67}=3(68)+2=204 + 2=206\) (since when \(n = 2\), it's the first term of our adjusted series, so the number of terms is \(68 - 2+1 = 67\)).

Step2: Apply the sum formula

Using the formula \(S=\frac{N}{2}(a_1 + a_N)\), where \(N = 67\), \(a_1 = 8\), \(a_N=206\). So \(S=\frac{67}{2}(8 + 206)=\frac{67}{2}\times214=67\times107\).

Step3: Calculate \(67\times107\)

\(67\times107=(70 - 3)\times(100 + 7)=70\times100+70\times7-3\times100 - 3\times7=7000+490 - 300 - 21=7000+490=7490;7490-300 = 7190;7190 - 21=7169\).

Answer:

7169