Find the sum (2 + 4 + 6 + 8 +… + 100).
Question:

Find the sum (2 + 4 + 6 + 8 +… + 100).

Solution:

It is required to find the sum of (2 + 4 + 6 + 8 +… 100).

Now, consider the series $(2+4+6+8+\ldots 100)$.

If we take a common factor of 2 from all the terms, then,

the series becomes,

2 (1 + 2 + 3 + 4 +… 50).

So, we need to find the sum of first 50 natural numbers.

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 first 50 natural numbers $=\frac{n(n+1)}{2}$

$=\frac{50(51)}{2}$

$=1275$

(2 + 4 + 6 + 8 +… 100) = 2 (1 + 2 + 3 + 4 +… 50)

= 2 x 1275 = 2550

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