Find the sum (41 + 42 + 43 + …. + 100).

It is required to find the sum (41 + 42 + 43 + …. + 100).

$(41+42+43+\ldots .+100)=$ Sum of integers starting from 1 to $100-$ Sum of integers starting from 1 to 40 .

Note:

Sum of first n natural numbers, 1 + 2 +3+…n,

$\sum_{k=1}^{n} k=\frac{n(n+1)}{2}$

From the above identities,

So, Sum of integers starting from 1 to 100 $=\frac{n(n+1)}{2}$

$=\frac{100(101)}{2}$

$=5050$

So, Sum of integers starting from 1 to 40 $=\frac{n(n+1)}{2}$

$=\frac{40(41)}{2}$

$=820$

$(41+42+43+\ldots+100)=$ Sum of integers starting from 1 to $100-$ Sum of integers starting from 1 to 40 .

$(41+42+43+\ldots+100)=5050-820=4230$