- #1
2sin54
- 109
- 1
Hi. I am learning mathematical induction on my own and I came across this problem:
Prove:
1*4 + 4*7 + 7*10 + ... + (3n - 2)(3n + 1) = n(n + 1)²2. The attempt at a solution
Quick test for n=1:
(3 -2)(3 + 1) = 1(1 + 1)²
4 = 4
Alright, so I rewrite this with, on the left side, after the '...' having two members:
1*4 + 4*7 + 7*10 + ... + (3n - 5)(3n - 2) + (3n - 2)(3n + 1) = n(n+1)²
Assume n = k and it stands for k:
1*4 + 4*7 + 7*10 + ... + (3k - 5)(3k - 2) + (3k - 2)(3k + 1) = k(k+1)²
Lets prove that it stands for n = k+1:
[1*4 + 4*7 + 7*10 + ... + (3k - 2)(3k + 1)] + (3k+1)(3k+4) = (k+1)(k+2)²
Now replace everything in brackets with k(k+1)² from the step above:
k(k+1)² + (3k+1)(3k+4) = (k+1)(k+2)²
k³ + 2k² + k + 9k² + 12k + 3k + 4 = (k+1)(k² + 4k + 4)
k³ + 11k² + 16k + 4 = k³ + 4k² + 4k + k² + 4k + 4 | -k³
11k² + 16k + 4 = 5k² + 8k + 4
6k² + 8k = 0Where's the mistake?
Homework Statement
Prove:
1*4 + 4*7 + 7*10 + ... + (3n - 2)(3n + 1) = n(n + 1)²2. The attempt at a solution
Quick test for n=1:
(3 -2)(3 + 1) = 1(1 + 1)²
4 = 4
Alright, so I rewrite this with, on the left side, after the '...' having two members:
1*4 + 4*7 + 7*10 + ... + (3n - 5)(3n - 2) + (3n - 2)(3n + 1) = n(n+1)²
Assume n = k and it stands for k:
1*4 + 4*7 + 7*10 + ... + (3k - 5)(3k - 2) + (3k - 2)(3k + 1) = k(k+1)²
Lets prove that it stands for n = k+1:
[1*4 + 4*7 + 7*10 + ... + (3k - 2)(3k + 1)] + (3k+1)(3k+4) = (k+1)(k+2)²
Now replace everything in brackets with k(k+1)² from the step above:
k(k+1)² + (3k+1)(3k+4) = (k+1)(k+2)²
k³ + 2k² + k + 9k² + 12k + 3k + 4 = (k+1)(k² + 4k + 4)
k³ + 11k² + 16k + 4 = k³ + 4k² + 4k + k² + 4k + 4 | -k³
11k² + 16k + 4 = 5k² + 8k + 4
6k² + 8k = 0Where's the mistake?