Question

Use mathematical induction to prove the statement is true for all positive integers n.

6 + 12 + 18 + ... + 6n = 3n(n + 1)

Answer 1

is this correct?

Step 1: 6 = 3(1)(1 + 1). true n=
Step 2: 16 + 12 + ... + 6k + 6(k+1) = [6 + 12 + ... + 6k] + 6(k + 1)
= 3k(k + 1) + 6(k + 1)
= 3(k + 1) [ k + 2 ]
= 3(k + 1)[(k + 1) + 1]
True for k+1
statement true for all positive integers

Answer 2

@ganeshie8

Answer 3

step 2 let
6 + 12 + 18 + ... + 6k = 3k(k + 1)


step 3 check for k+1

Answer 4

so wanna prove
6 + 12 + 18 + ... + 6(k+1) = 3(k+1)(k + 2)

Answer 5

mine is wrong?

Answer 6

ur is not wrong , but when u have induction u have t set the steps in order

steps for induction
1_ check for n=1
2_ assume its correct for n
3_ show its correct for n+1

Answer 7

so step3 would be like
show for k+1
6 + 12 + 18 + ... + 6(k+1) = 3(k+1)(k + 2)
6 + 12 + 18 + ... + 6(k+1)= 6 + 12 + 18 + ... + 6 k +6=[6 + 12 + 18 + ... + 6 k] +6
induction step :-
6 + 12 + 18 + ... + 6k = 3k(k + 1)

[6 + 12 + 18 + ... + 6 k] +6 =3k(k + 1) +6 = 3(k(k+1) +2 ) =3(k+1)(k + 2)