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)$31 + 32 +… + 50$

To do:

We have to find the values of the given series.

Solution:

We observe that,

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

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

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

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

Therefore,

(i) $1+2+3+4+5+\ldots . .+50=\frac{1}{2}\{50 \times(50+1)\}$

$=\frac{1}{2} \times 50 \times 51$

$=1275$

(ii) $(1+2+3+4+\ldots . .+50)-(1+2+3+4+\ldots . .+30)=\frac{1}{2}\{50 \times(50+1)\}-\frac{1}{2}\{30 \times(30+1)\}$

$=\frac{1}{2} \times 50 \times 51-\frac{1}{2} \times 30 \times 31$

$=1275-465$

$=810$