Question

Use mathematical induction to prove the following statement:

Answer 1

Answer:

Step-by-step explanation:

To prove that the sum of the first n even +ve integers is:

$\mathsf{2+4+6+8+ . . . +2n = n^2+ n }$

By using mathematical induction;

For n = 1, we get:

2n = 2 × 1 = 2

2 = 1² + 1 ----- (1)

∴ the outcome is true if n = 1

However, let assume that the result is also true for n = k

Now, $\mathsf{2+4+6+8+. . .+2k = k^2 + k --- (2)}$

$\mathsf{2+4+6+8+. . .+2k+2(k+1)}$

we can now say:

$\mathsf{= (k^2 + k) + 2(k + 1)} \\ \\ \mathsf{= k^2 + k + 2k + 1}$

$\mathsf{= (k^2 + 2k + 1) + (k + 1)}$

$\mathsf{= (k + 1)^2 + (k + 1)}$

∴

$\mathsf{2 + 4 + 6 + 8 + . . . + 2k + 2(k + 1) = (k + 1)^2 + (k + 1)}$

Thus, the result is true for n = m+1, hence we can posit that the result is also true for each value of n.

As such $\mathsf{2+4+6+8+. . .+2n = n^2 + n }$