Question

∑62 (4n+7) Find the numerical answer to the summation given below.

Answer 1

To evaluate the summation \(\sum_{n=1}^{62} (4n + 7)\), we can split the summation into two parts:

\[
\sum_{n=1}^{62} (4n + 7) = \sum_{n=1}^{62} 4n + \sum_{n=1}^{62} 7
\]

Now, let's calculate each part.

### Calculating \(\sum_{n=1}^{62} 4n\):

This can be factored as follows:

\[
\sum_{n=1}^{62} 4n = 4 \sum_{n=1}^{62} n
\]

The formula for the sum of the first \(N\) natural numbers is:

\[
\sum_{n=1}^{N} n = \frac{N(N + 1)}{2}
\]

For \(N = 62\):

\[
\sum_{n=1}^{62} n = \frac{62 \times 63}{2} = \frac{3906}{2} = 1953
\]

Using this result, we can find \(\sum_{n=1}^{62} 4n\):

\[
\sum_{n=1}^{62} 4n = 4 \times 1953 = 7812
\]

### Calculating \(\sum_{n=1}^{62} 7\):

This is straightforward since \(7\) is a constant:

\[
\sum_{n=1}^{62} 7 = 7 \times 62 = 434
\]

### Combining the results:

Now we add both results together:

\[
\sum_{n=1}^{62} (4n + 7) = \sum_{n=1}^{62} 4n + \sum_{n=1}^{62} 7 = 7812 + 434 = 8246
\]

Thus, the numerical answer to the summation is:

\[
\boxed{8246}
\]