Evaluate the Summation sum from 12 to 73 of n

Solution:

Step 1:

Reformulate the given summation to start from 1 by separating it into two parts.

$\sum _{n=12}^{73}n=\sum _{n=1}^{73}n-\sum _{n=1}^{11}n$

Step 2:

Calculate the summation from 1 to 73.

Step 2.1:

Use the arithmetic series sum formula:

$\sum _{k=1}^{n}k=\frac{n(n+1)}{2}$

Step 2.2:

Insert 73 into the formula:

$\frac{73(73+1)}{2}$

Step 2.3:

Perform the arithmetic operations.

Step 2.3.1:

Add together 73 and 1.

$\frac{73\times 74}{2}$

Step 2.3.2:

Multiply 73 by 74.

$\frac{5402}{2}$

Step 2.3.3:

Divide 5402 by 2.

$2701$

Step 3:

Calculate the summation from 1 to 11.

Step 3.1:

Use the same arithmetic series sum formula:

$\sum _{k=1}^{n}k=\frac{n(n+1)}{2}$

Step 3.2:

Insert 11 into the formula:

$\frac{11(11+1)}{2}$

Step 3.3:

Perform the arithmetic operations.

Step 3.3.1:

Add together 11 and 1.

$\frac{11\times 12}{2}$

Step 3.3.2:

Multiply 11 by 12.

$\frac{132}{2}$

Step 3.3.3:

Divide 132 by 2.

$66$

Step 4:

Combine the calculated values from the two summations.

$2701-66$

Step 5:

Complete the subtraction to find the final result.

$2635$

Knowledge Notes:

The problem involves evaluating the summation of a sequence of integers from 12 to 73. The solution uses the formula for the sum of the first $n$ natural numbers, which is an arithmetic series. The formula is:

$$\sum _{k=1}^{n}k=\frac{n(n+1)}{2}$$

This formula is derived from the observation that the sum of a series of natural numbers can be paired off into terms with the same total (first and last, second and second-to-last, etc.), and there are $n/2$ such pairs if $n$ is even (or $(n+1)/2$ pairs if $n$ is odd). The process involves breaking the original summation into two parts: one from 1 to 73 and another from 1 to 11, then evaluating each using the formula and subtracting the second from the first to get the final result. This method is based on the principle of superposition, which allows for the separation and independent evaluation of each summation.