Question

Solve using principle of mathematical induction, prove that for every n >= 1, 7 + 13 + 19 + . . . + (6n + 1) = n (3n +4)

Answer 1

Answer:

Let P(n) denote the statement :

$\text{For n\geq1 , 7+13+19+.......+(6n+1)=n(3n+4)}$

Put n=1,

L.H.S=7

R.H.S=3(1)+4=7

$\therefore\:$ P(1) is true

Assume that P(k) is true

That is

7+13+19+.......+(6k+1)=k(3k+4)  is true

To prove: P(k+1) is true

That is , to prove

$7+13+19+.......+(6k+7)=(k+1)(3k+7)$  is true

Now, consider

7+13+19+.......+(6k+7)

$=7+13+19+.......+(6k+1)+(6k+7)$

$=[7+13+19+.......+(6k+1)]+(6k+7)$

$=[k(3k+4)]+(6k+7)$

$=3k^2+4k+6k+7$

$=3k^2+10k+7$

$=3k^2+3k+7k+7$

$=3k(k+1)+7(k+1)$

$=(k+1)(3k+7)$

Thus, P(k+1) is true

Hence by mathematical induction P(n) is true for all natural numbers