# Find the sum of the \(\sum_{i=5}^{13}3i + 2\)

**Solution:**

\(\sum_{i=5}^{13}3i + 2=\sum_{i=1}^{13}3i + 2-\sum_{i=1}^{4}3i + 2\)

First let us consider \(\sum_{i=1}^{13}3i + 2\)

To fit the summation rules let us split it into smaller summations

\(\sum_{i=1}^{13}3i + 2=3\sum_{i=1}^{13}i+\sum_{i=1}^{13}2\)

For summation of a polynomial with degree 1

\(\sum_{i=1}^{n}i=\frac{n(n+1)}{2}\)

By substituting the values

\((3)(\frac{13(13+1)}{2})\)

On further simplification

= 3 (13 × 14)/2

= 273

The summation of a constant formula is

\(\sum_{i=1}^{n}c=cn\)

By substituting the values

= (2) (13)

= 26

So we get

\(\sum_{i=1}^{13}3i + 2=3\sum_{i=1}^{13}i+\sum_{i=1}^{13}2\)

= 273 + 26

= 299

To fit the summation rules let us split it into smaller summations

\(\sum_{i=1}^{4}3i+2=3\sum_{i=1}^{4}i+\sum_{i=1}^{4}2\)

For summation of a polynomial with degree 1

\(\sum_{i=1}^{n}i=\frac{n(n+1)}{2}\)

By substituting the values

\((3)(\frac{4(4+1)}{2})\)

On further simplification

= 3 (4 × 5)/2

= 30

The summation of a constant formula is

\(\sum_{i=1}^{n}c=cn\)

By substituting the values

= (2) (4)

= 8

So we get

\(\sum_{i=1}^{4}3i+2=3\sum_{i=1}^{4}i+\sum_{i=1}^{4}2\)

= 30 + 8

= 38

Here

\(\sum_{i=5}^{13}3i + 2=\sum_{i=1}^{13}3i + 2-\sum_{i=1}^{4}3i + 2\)

= 299 - 38

= 261

Therefore, the sum is 261.

## Find the sum of the \(\sum_{i=5}^{13}3i + 2\)

**Summary:**

The sum of \(\sum_{i=5}^{13}3i + 2\) is 261.

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