Evaluate the \(\sum_{n=1}^{14}\) 3n + 2

Solution:

Given, \(\sum_{n=1}^{14}\) 3n + 2

We have to evaluate the given summation.

When n = 1, 3(1) + 2 = 3 + 2 = 5

When n = 2, 3(2) + 2 = 6 + 2 = 8

When n = 3, 3(3) + 2 = 9 + 2 = 11

When n = 4, 3(4) + 2 = 12 + 2 = 14

When n = 5, 3(5) + 2 = 15 + 2 = 17

From the sequence 5, 8, 11, 14, 17,....

The sequence is in arithmetic progression.

First term a = 5

Common difference d = 8 - 5 = 3

We have to find the sum of the given arithmetic series upto 14 terms.

The sum of the n terms of arithmetic sequence is given by

\(s_{n}=\frac{n}{2}(a+l)\)

Where, n = number of terms

a = first term

l = last term

Here, last term l = 3(14) + 2 = 44

Now, a = 5, l = 44, n = 14

\(s_{14}=\frac{14}{2}(5+44)\)

\(s_{14}=7(49)\)

\(s_{14}=343\)

Therefore, \(\sum_{n=1}^{14}\) 3n + 2 is 343.

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Evaluate the \(\sum_{n=1}^{14}\) 3n + 2

Summary:

On evaluating \(\sum_{n=1}^{14}\) 3n + 2, the solution is 343.