Question

mathematical induction. can anyone please help me ?

Answer 1

1) works for the 1st term (show this)
2) assume it is true for any number, k
3)prove it is true for the next term, k+1, by proving that:

Sk + a(k+1) = S(k+1)

example
Prove the following using Mathematical Induction.

Sn = 6 + 12 + 18 + … + 6n = 3n(n+1)
1. Prove that it is true for S1.

S1 = 3(1)(1+1) = 3(2) = 6

2. Prove that it is true for Sk.

Sk = 6 + 12 + 18 + … + 6k = 3k(k+1)

3. Prove that it is true for Sk+1.

Sk+1 = 6 + 12 + 18 + … + 6k + 6(k + 1) = 3(k+1)(k+1+1)

Sk+1 = 6 + 12 + 18 + … + 6k + 6(k + 1) = 3(k+1)(k+1+1)

Sk+1 = 6 + 12 + 18 + … + 6k + 6(k + 1) = 3(k+1)(k+2)


Sk+1 = 6 + 12 + 18 + … + 6k + 6(k + 1) = 3(k+1)(k+1+1)

Sk+1 = 6 + 12 + 18 + … + 6k + 6(k + 1) = 3(k+1)(k+2)

6 + 12 + 18 + … + 6k + 6(k + 1) = 3(k+1)(k+1+1)

6 + 12 + 18 + … + 6k + 6(k + 1) = 3(k+1)(k+2)

3k(k+1) + 6(k + 1) = 3(k+1)(k+2)

Sk+1 = 6 + 12 + 18 + … + 6k + 6(k + 1) = 3(k+1)(k+1+1)

Sk+1 = 6 + 12 + 18 + … + 6k + 6(k + 1) = 3(k+1)(k+2)

6 + 12 + 18 + … + 6k + 6(k + 1) = 3(k+1)(k+1+1)

6 + 12 + 18 + … + 6k + 6(k + 1) = 3(k+1)(k+2)

3k(k+1) + 6(k + 1) = 3(k+1)(k+2)

3k2 + 3k + 6k + 6 = 3(k+1)(k+2)


Sk+1 = 6 + 12 + 18 + … + 6k + 6(k + 1) = 3(k+1)(k+1+1)

Sk+1 = 6 + 12 + 18 + … + 6k + 6(k + 1) = 3(k+1)(k+2)

6 + 12 + 18 + … + 6k + 6(k + 1) = 3(k+1)(k+1+1)

6 + 12 + 18 + … + 6k + 6(k + 1) = 3(k+1)(k+2)

3k(k+1) + 6(k + 1) = 3(k+1)(k+2)

3k2 + 3k + 6k + 6 = 3(k+1)(k+2)

3k2 + 9k + 6 = 3(k+1)(k+2)

Sk+1 = 6 + 12 + 18 + … + 6k + 6(k + 1) = 3(k+1)(k+1+1)

Sk+1 = 6 + 12 + 18 + … + 6k + 6(k + 1) = 3(k+1)(k+2)

6 + 12 + 18 + … + 6k + 6(k + 1) = 3(k+1)(k+1+1)

6 + 12 + 18 + … + 6k + 6(k + 1) = 3(k+1)(k+2)

3k(k+1) + 6(k + 1) = 3(k+1)(k+2)

3k2 + 3k + 6k + 6 = 3(k+1)(k+2)

3k2 + 9k + 6 = 3(k+1)(k+2)

3(k2 + 3k + 2) = 3(k+1)(k+2)

3(k+1)(k+2) = 3(k+1)(k+2)