Proof by mathematical induction: 1^2+42+7^2+10^2+13^2+...+(6n+1)^2=(2n+1)(12n^2+9n+ 1)

[rolevel]

Proof by mathematical induction: 1^2+42+7^2+10^2+13^2+...+(6n+1)^2=(2n+1)(12n^2+9n+ 1)

I am trying to solve next problem:
1²+4²+7²+10²+13² + ... + (6n+1)² = (2n+1)(12n² +9n + 1) for n>=1

I found myself stuck at point, where i need to replace k by k+1 and finish proof. Usually when you have something like this: 1+2+3+4+5+...+n+(n+1) = (n+1)(n+2)/2, bold will be replaced with n(n+1)/2, so you can solve it then. But it doesn't seem to work out with problem i described above. This is where i stuck:

1²+4²+7²+10²+13² + ... + (6k+1)² +(6k+7)² = (2k+3)(12k²+33k+22)
(2n+1)(12n² +9n + 1) + (6k+7)

Can anyone please explain me what i am doing wrong?

 

[Deleted member 4993]

I am trying to solve next problem:
1²+4²+7²+10²+13² + ... + (6n+1)² = (2n+1)(12n² +9n + 1) for n>=1

I found myself stuck at point, where i need to replace k by k+1 and finish proof. Usually when you have something like this: 1+2+3+4+5+...+n+(n+1) = (n+1)(n+2)/2, bold will be replaced with n(n+1)/2, so you can solve it then. But it doesn't seem to work out with problem i described above. This is where i stuck:

1²+4²+7²+10²+13² + ... + (6k+1)² +(6k+7)² = (2k+3)(12k²+33k+22)
(2n+1)(12n² +9n + 1) + (6k+7)

Can anyone please explain me what i am doing wrong?

There is something very wrong with your problem statement!

That should be: (3*n-2)²

The expression for summed series is not correct for even n=1.

 

[rolevel]

There is something very wrong with your problem statement!

That should be: (3*n-2)²

The expression for summed series is not correct for even n=1.

Base on what part b states, it suppose to be true.

After i post the thread i found that example in my book, but i still not quite understand the way it works.

 

[Steven G]

Base on what part b states, it suppose to be true.
View attachment 10298
After i post the thread i found that example in my book, but i still not quite understand the way it works.
View attachment 10300

Whether you found the problem in your textbook or the garbage can (where it belongs) does not matter, The term (6n+1)² is the general term (or at least should be). That is when n=1 the 1st term which should be (6*1+1)^2 which is 49 BUT the first term is 1 NOT 49. The second term should be (6*2+1)^2 = 169 which is NOT 4. .... Conclusion is that the general term is NOT (6n+1)². Since the bases of consectutive terms all differ by 3, the general term for the base will be in the form 3n + k (k is some number to be computed) and certainly NOT in the form of 6n+k!!

Another words the theorem is NOT true!

 

[JeffM]

I am trying to solve next problem:
1²+4²+7²+10²+13² + ... + (6n+1)² = (2n+1)(12n² +9n + 1) for n>=1

I found myself stuck at point, where i need to replace k by k+1 and finish proof. Usually when you have something like this: 1+2+3+4+5+...+n+(n+1) = (n+1)(n+2)/2, bold will be replaced with n(n+1)/2, so you can solve it then. But it doesn't seem to work out with problem i described above. This is where i stuck:

1²+4²+7²+10²+13² + ... + (6k+1)² +(6k+7)² = (2k+3)(12k²+33k+22)
(2n+1)(12n² +9n + 1) + (6k+7)

Can anyone please explain me what i am doing wrong?

The most likely situations are that (1) there is an error in the book, or (2) the OP did not transcribe the problem correctly.

\(\displaystyle \text {PROVE: } \displaystyle n \in \mathbb Z^+ \implies (6n + 1)^2 + \sum_{j=1}^{n+1} (3j - 2)^2 \le (2n + 1)(12n^2 + 9n + 1) \implies\)

\(\displaystyle \displaystyle (6n + 1)^2 + \sum_{j=1}^{n+1} (3j - 2)^2 \le 24n^3 + 30n^2 + 11n + 1.\)

\(\displaystyle \displaystyle n = 1 \implies (6 * 1 + 1)^2 + \sum_{j=1}^{1+1} (3j - 2)^2 = 7^2 + 1^2 + 4^2 = 66 =\)

\(\displaystyle \displaystyle 24 + 30 + 11 + 1 \le 24 * 1^3 + 30 * 1^2 + 11 * 1 + 1.\)

\(\displaystyle \displaystyle n = 1 \implies (6n + 1)^2 + \sum_{j=1}^{n+1} (3j - 2)^2 \le 24n^3 + 30n^2 + 11n + 1.\)

\(\displaystyle \therefore \exists \ \mathbb K \text { such that } \mathbb K \text { is not empty, and } k \in \mathbb K \implies\)

\(\displaystyle \displaystyle k \in \mathbb Z^+ \text { and } (6k + 1)^2 + \sum_{j=1}^{k+1} (3j - 2)^2 \le 24k^3 + 30k^2 + 11k + 1.\)

\(\displaystyle \{ 6 * ( k + 1 ) + 1\}^2\) \(\displaystyle \displaystyle + \sum_{j=1}^{ ( k + 1 ) + 1 } (3j - 2)^2 =\)

\(\displaystyle \displaystyle (6k + 6 + 1)^2 + \sum_{j=1}^{k+2} (3j - 2)^2 = (6k + 7)^2 + \{3(k + 2) - 2\}^2 + \sum_{j=1}^{k+1} (3j - 2)^2 =\)

\(\displaystyle \displaystyle (6k + 7)^2 + (3k + 6 - 2)^2 + \sum_{j=1}^{k+1} (3j - 2)^2 = (6k + 7)^2 + (3k + 4)^2 + \sum_{j=1}^{k+1} (3j - 2)^2 =\)

\(\displaystyle \displaystyle 36k^2 + 84k + 49 + 9k^2 + 24k + 16 + \sum_{j=1}^{k+1} (3j - 2)^2 =\)

\(\displaystyle \displaystyle (36k^2 + 12k + 1) + 9k^2 + 96k + 64 + \sum_{j=1}^{k+1} (3j - 2)^2 =\)

\(\displaystyle \displaystyle 9k^2 + 96k + 64 + \left ( (6k + 1)^2 + \sum_{j=1}^{k+1} (3j - 2)^2 \right ).\)

\(\displaystyle \therefore \displaystyle \{6(k + 1) + 1\}^2 + \sum_{j=1}^{(k+1)+1} (3j - 2)^2 = 9k^2 + 96k + 64 + \left ( (6k + 1)^2 + \sum_{j=1}^{k+1} (3j - 2)^2 \right ).\)

\(\displaystyle \{2(k + 1) + 1\}\{12(k + 1)^2 + 9(k + 1) + 1\} = (2k + 3)(12k^2 + 33k + 22) =\)

\(\displaystyle 24k^3 + 66k^2 + 44k + 36k^2 + 99k + 66 = 24k^3 +102k^2 + 143k + 66 =\)

\(\displaystyle 72k^2 + 132k + 65 + (24k^3 + 30k^2 + 11k + 1) .\)

\(\displaystyle 72k^2 + 132k + 65 - (9k^2 + 96k + 64) = 63k^2 + 34k + 1 > 0 \ \because \ k \in \mathbb Z^+.\)

\(\displaystyle \text {Or } 9k^2 + 96k + 64 < 72k^2 + 132k + 65.\)

\(\displaystyle \displaystyle (6k + 1)^2 + \sum_{j=1}^{k+1} (3j - 2)^2 \le 24k^3 + 30k^2 + 11k + 1.\)

\(\displaystyle \therefore \displaystyle 9k^2 + 96k + 64 + (6k + 1)^2 + \sum_{j=1}^{k+1} (3j - 2)^2 \le \\

72k^2 + 132k + 65 + (24k^3 + 30k^2 + 11k + 1) \implies\)

\(\displaystyle \displaystyle \{6(k + 1) + 1\}^2 + \sum_{j=1}^{(k+1)+1} (3j - 2)^2 \le \\

\{2(k + 1) + 1\}\{12(k + 1)^2 + 9(k + 1) + 1\}.\)

\(\displaystyle \text {Q.E.D.}\)

Dumb question (if that even was the question). I think it may have been because the equality holds for n = 1 so it does not appear to be utterly fanciful. A ton of ugly algebra to demonstrate a basic technique.