Observe the following pattern \[1 = \frac{1}{2}\left\{ 1 \times \left( 1 + 1 \right) \right\}\]

\[ 1 + 2 = \frac{1}{2}\left\{ 2 \times \left( 2 + 1 \right) \right\}\]

\[ 1 + 2 + 3 = \frac{1}{2}\left\{ 3 \times \left( 3 + 1 \right) \right\}\]

\[1 + 2 + 3 + 4 = \frac{1}{2}\left\{ 4 \times \left( 4 + 1 \right) \right\}\]and find the values of following:

31 + 32 + ... + 50

#### Solution

Observing the three numbers for right hand side of the equalities:

The first equality, whose biggest number on the LHS is 1, has 1, 1 and 1 as the three numbers.

The second equality, whose biggest number on the LHS is 2, has 2, 2 and 1 as the three numbers.

The third equality, whose biggest number on the LHS is 3, has 3, 3 and 1 as the three numbers.

The fourth equality, whose biggest number on the LHS is 4, has 4, 4 and 1 as the three numbers.

Hence, if the biggest number on the LHS is *n*, the three numbers on the RHS will be *n*, *n*and 1.

Using this property, we can calculate the sums for (i) and (ii) as follows:

The sum can be expressed as the difference of the two sums as follows:

\[31 + 32 + . . . . . + 50 = (1 + 2 + 3 + . . . . . . + 50) - ( 1 + 2 + 3 + . . . . . . + 30)\]

The result of the first bracket is exactly the same as in part (i).

\[1 + 2 + . . . . + 50 = 1275\]

Then, the second bracket:

\[1 + 2 + . . . . . . + 30 = \frac{1}{2}\left( 30 \times \left( 30 + 1 \right) \right)\]

Finally, we have:

\[31 + 32 + . . . . + 50 = 1275 - 465 = 810\]