Question

Prove by the principle of mathematical induction that for all n belongs to N:
​

Answer 1

Step-by-step explanation:

I hope it will help you to understand the solution

Answer 2

Answer:

hello there

Let's prove it first for

P(1)

$\frac{1}{2} \times 1(3 \times (1) - 1)$

$\frac{1}{2} \times 1 \times 2 = 1$

hence it's true for P(1)

let P(k) be :

1+ 4+ 7 +....+(3k - 2) =

$\frac{1}{2} k(3k - 1)$

_____(equation 1)

now

We have to prove for p(k+1)

1+4+7+.....+(3k-2)+[3(k+1)-2]=

$\frac{1}{2} (k + 1)[3(k + 1) - 1]$

____________________________________

now from equation 1 we have

$\frac{1}{2} k( 3k - 1) + (3k + 3 - 2)$

=>$\frac{1}{2}3k ^{2} - k \: + (3k + 1)$

=>$\frac{ {3k }^{2} - k + 6k + 2 }{2} =>$

=>$\frac{ {3k}^{2} + 2k + 3k + 2}{2}$

=>$\frac{3k(k + 1) + 2(k + 1)}{2}$

=>$\frac{(3k + 2)(k + 1)}{2}$

=>$\frac{1}{2} (k + 1)[3(k + 1) - 1]$

Thank you ｡◕‿◕｡