Question

Is this correct?

Pre Cal Math

Answer 1

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

10 + 30 + 60 + ... + 10n = 5n(n + 1)

Answer 2

Part 1: 10(1) = 5(1)(1+1) ?
10 = 5 * 2
10 = 10 OK

Part 2: 10+30+... + 10(k) = 5(k)(k + 1)

Part 3: 10+30+... + 10k + 10(k+1) = 5k(k + 1) + 10(k+1)
Right side must equal 5(k + 1)((k + 1) + 1) for it to be true

5k(k + 1) + 10(k + 1)
(k + 1)(5k + 10)
5(k + 1)(k + 2)
5(k + 1)((k + 1) + 1) OK

Answer 3

for all positive integers, so it must be true for n=1 to move ahead

10(1) = 5(1)(1 + 1)
10 = 10 is true

lets assume its true for some generic n=k

P(k) = 10+30+... + 10k = 5k (k + 1)

P(k+1) = 10+30+... + 10k + 10(k+1) = P(k) + 10(k+1)

lets add 10(k+1) to both sides

10+30+... + 10k + 10(k+1) = 5k (k + 1) + 10(k+1)

can we simplify the right side to the format: 5(k+1)(k+1+1) ??

Answer 4

5k (k + 1) + 10(k+1)

5 [k (k + 1) + 2k+1 ]

5 [k^2 + k + 2k + 1 ]

5 [k^2 + 3k + 1 ]

5 (k+1) (....) hmmm, did i make an error?

Answer 5

second line, 2k+2 yeah

Answer 6

5k (k + 1) + 10(k+1)

5 [k (k + 1) + 2(k+1) ]

5 [k^2 + k + 2k + 2 ]

5 [k^2 + 3k + 2 ]

5 (k+1) (k+2), good

Answer 7

did i make a mistake?

Answer 8

since we know its good for k and k+1; we know its good for k=1, it has to be good for k=2,3,4,5,.... as well

Answer 9

nah, i think you did fine