Observe the following pattern:
$2^2 – 1^2 = 2 + 1$
$3^2 – 2^2 = 3 + 2$
$4^2 – 3^2 = 4 + 3$
$5^2 – 4^2 = 5 + 4$
Find the value of
(i) $100^2 – 99^2$
(ii) $111^2 – 109^2$
(iii) $99^2 – 96^2$

To do:

We have to write the value of given expressions.

Solution:

We can observe that,

$2^2 – 1^2 = 2 + 1$

$3^2 – 2^2 = 3 + 2$

$4^2 – 3^2 = 4 + 3$

$5^2 – 4^2 = 5 + 4$

This implies,

(i) $m^2-n^2=m+n$

Therefore,

$100^2 – 99^2=100+99$.

(ii) $m^2-n^2=m+n$

$111^2 – 109^2$ can be written as,

$111^2 – 109^2=111^2-110^2+110^2-109^2$

Therefore,

$111^2 – 109^2=(111^2-110^2)+(110^2-109^2)$

$=(111+110)+(110+109)$

$=440$ 

(iii) $m^2-n^2=m+n$

$99^2 – 96^2$ can be written as,

$99^2 – 96^2=99^2-98^2+98^2-97^2+97^2-96^2$

Therefore,

$99^2 – 96^2=(99^2-98^2)+(98^2-97^2)+(97^2-96^2)$

$=(99+98)+(98+97)+(97+96)$

$=585$