**Question:**

Observe the following pattern

$1=\frac{1}{2}\{1 \times(1+1)\}$

$1+2=\frac{1}{2}\{2 \times(2+1)\}$

$1+2+3=\frac{1}{2}\{3 \times(3+1)\}$

$1+2+3+4=\frac{1}{2}\{4 \times(4+1)\}$

and find the values of each of the following:

(i) 1 + 2 + 3 + 4 + 5 + ... + 50

(ii) 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:

(i) $1+2+3+\ldots \ldots \ldots+50=\frac{1}{2} \times 50 \times(50+1)=1275$

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

$31+32+\ldots+50=(1+2+3+\ldots \ldots+50)-(1+2+3+\ldots \ldots+30)$

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

$1+2+\ldots+50=1275$

Then, the second bracket:

$1+2+\ldots \ldots+30=\frac{1}{2}(30 \times(30+1))=465$

Finally, we have:

$31+32+\ldots+50=1275-465=810$