Question

sum of all the first n terms of even natural number is

Answer 1

Solution:

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Given & To Find:

The sum of first n even numbers,.

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The first even number = 2 (excluding 0)

d = 2,/ ( common difference between 2 even numbers)

We know that,

$S_n= \frac{n}{2}(2a + (n-1)d)$

Substituting the known values,

We get,

=> $S_n = \frac{n}{2}(2(2)+ (n-1)(2))$

=> $S_n= \frac{n}{2}(2(2 + (n-1)))$

=> $S_n = n (2 + n-1)$

=> ∴ $S_n = n (n+1) =n^2 +n$

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                        Hope it Helps!!

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Answer 2

Step-by-step explanation:

first term : 2

common difference : 4-2=2

Sn = n/2[2t1+(n-1)d]

= n/2(2x2+(n-1)2)

= n/2(4+2n-2)

= n/2(2n+2) {when 2 is cancelled }

= 2n^2