# Observe the following pattern

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 nn 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:

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