Question

Observe the following pattern
$$\begin{gathered} 1=\frac{1}{2}\left\{1\times \left(1+1\right)\right\} \\ 1+2=\frac{1}{2}\left\{2\times \left(2+1\right)\right\} \\ 1+2+3=\frac{1}{2}\left\{3\times \left(3+1\right)\right\} \\ 1+2+3+4=\frac{1}{2}\left\{4\times \left(4+1\right)\right\} \end{gathered}$$
and find the values of each of the following:
(i) 1 + 2 + 3 + 4 + 5 + ... + 50
(ii) 31 + 32 + ... + 50

Answer 1

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:

$\left(\mathrm{i}\right)1+2+3+........+50=\frac{1}{2}\times 50\times (50+1)=1275$

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

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

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

$1+2+....+50=1275$

Then, the second bracket:
$1+2+......+30=\frac{1}{2}\left(30\times \left(30+1\right)\right)$ = 465
Finally, we have:
$31+32+....+50=1275-465=810$