**Question:**

Observe the following pattern

$1^{2}=\frac{1}{6}[1 \times(1+1) \times(2 \times 1+1)]$

$1^{2}+2^{2}=\frac{1}{6}[2 \times(2+1) \times(2 \times 2+1)]$

$1^{2}+2^{2}+3^{2}=\frac{1}{6}[3 \times(3+1) \times(2 \times 3+1)]$

$1^{2}+2^{2}+3^{2}+4^{2}=\frac{1}{6}[4 \times(4+1) \times(2 \times 4+1)]$

and find the values of each of the following:

(i) 12 + 22 + 32 + 42 + ... + 102

(ii) 52 + 62 + 72 + 82 + 92 + 102 + 112 + 122

**Solution:**

Observing the six numbers on the RHS of the equalities:

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

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

The third equality, whose biggest number on the LHS is 3, has 3, 3, 1, 2, 3 and 1 as the six numbers.

The fourth equality, whose biggest number on the LHS is 4, has numbers 4, 4, 1, 2, 4 and 1 as the six numbers.

Note that the fourth number on the RHS is always 2 and the sixth number is always 1. The remaining numbers are equal to the biggest number on the LHS.

Hence, if the biggest number on the LHS is *n*, the six numbers on the RHS would be *n*, *n*, 1, 2, *n* and 1.

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

(i) $1^{2}+2^{2}+\ldots \ldots+10^{2}=\frac{1}{6} \times[10 \times(10+1) \times(2 \times 10+1)]$

$=\frac{1}{6} \times[10 \times 11 \times 12]$

$=385$

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

$5^{2}+6^{2}+\ldots \ldots+12^{2}=\left(1^{2}+2^{2}+\ldots \ldots+12^{2}\right)-\left(1^{2}+2^{2}+\ldots \ldots+4^{2}\right)$

The sum of the first bracket on the RHS:

$1^{2}+2^{2}+\ldots \ldots+12^{2}=\frac{1}{6}[12 \times(12+1) \times(2 \times 12+1)]$

= 650

The second bracket is:

$1^{2}+2^{2}+\ldots \ldots+4^{2}$

$=\frac{1}{6} \times[4 \times(4+1) \times(2 \times 4+1)]$

$=\frac{1}{6} \times 4 \times 5 \times 9=30$

Finally, the wanted sum is:

$5^{2}+6^{2}+\ldots \ldots+12^{2}$

$=\left(1^{2}+2^{2}+\ldots \ldots+12^{2}\right)-\left(1^{2}+2^{2}+\ldots \ldots+12^{2}\right)$

$=650-30=620$