Question

$\star\mathrm{solve \: this \: with \: explanation}$
​

Answer 1

$\sf\large\underline\green{AnSWEr}$

x1 = 3

x(k + 1) = xk + 2k + 3, for k ≥ 1 .

Use induction to prove that

xn = n (n + 2) for n ≥ 1 .

base case:

x1 = 3; 1 (1 + 2) = 3 , v (correct)

2.

Suppose that xk = k (k + 2), for k ≥ 1 .

Then

x(k + 1)

= k (k + 2) + 2k + 3

= k^2 + 4k + 3

= (k + 1) (k + 3)

= (k + 1) ( (k + 1) + 2) .

I..e., it follows that the proposed formula for xn holds also for n = k + 1 .

So xn = n (n + 2) for all n ≥ 1 .

Answer 2

$\sf\large\underline\green{AnSWEr}$

x1 = 3

x(k + 1) = xk + 2k + 3, for k ≥ 1 .

Use induction to prove that

xn = n (n + 2) for n ≥ 1 .

base case:

x1 = 3; 1 (1 + 2) = 3 , v (correct)

2.

Suppose that xk = k (k + 2), for k ≥ 1 .

Then

x(k + 1)

= k (k + 2) + 2k + 3

= k^2 + 4k + 3

= (k + 1) (k + 3)

= (k + 1) ( (k + 1) + 2) .

I..e., it follows that the proposed formula for xn holds also for n = k + 1 .

So xn = n (n + 2) for all n ≥ 1 .

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