Question

What is the sum of first k+1 even numbers

Answer 1

Answer:

$Sum = (k+1)(k+2)$

Step-by-step explanation:

Given

k + 1 even numbers

Required

Their sum

This question will be solved using the sum of nth term of an AP

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

Where

a = the first even number

$a = 2$

n = number of terms

$n =k +1$

d = difference between consecutive even numbers

$d = 2$

So, the expression becomes:

$Sum = \frac{k+1}{2}(2 * 2 + (k + 1 - 1) * 2)$

$Sum = \frac{k+1}{2}(2 * 2 + (k) * 2)$

$Sum = \frac{k+1}{2}(4 + 2k)$

$Sum = \frac{(k+1)(4 + 2k)}{2}$

Factorize 4 + 2k

$Sum = \frac{(k+1)*2*(2 + k)}{2}$

$Sum = (k+1)*(2 + k)$

$Sum = (k+1)*(k+2)$

$Sum = (k+1)(k+2)$