Problem w/ induction? How does 1/2k(6k^2 + 3k − 1) + [3(k+1)-1]^2 become...?

[manishpamnani169]

We have a question which says:-

Prove for any natural number n that:-

2² + 5² + 8² + ··· + (3n − 1)² = 1/2n(6n² + 3n − 1)

and here is the solution:-
P(n): 2²+5²+8²+ . . .+ (3n-1)² = 1/2n(6n² + 3n − 1)

For 'n' = 1, P(1) = 2² = 1/2(1)[6(1)²+3(1) -1 ]
4 = 1/2(1)[6(1)²+3(1)-1]
4 = 1/2 * 8
4 = 4
LHS = RHS
So, P(1) is true.

Assume that P(k) is true for some integer 'k'
P(k):- 2²+5²+8²+ . . .+ (3k-1)² = 1/2k(6k² + 3k − 1)------------ eq(1)

We shall now prove that P(k+1) is true. Now we have:-
2²+5²+8²+ . . .+ (3k-1)² +[3(k+1)-1]² = 1/2k(6k² + 3k − 1) +[3(k+1) -1]²

using LHS,
=> 2²+5²+8²+ . . .+ (3k-1)² +[3(k+1)-1]²
=> 1/2k(6k² + 3k − 1) + [3(k+1)-1]² ---- Using eq(1)
=> 1/2k(6k²+ 3k − 1) + 9(k+1)² - 6(k+1) + 1

Now how 1/2k(6k² + 3k − 1) + [3(k+1)-1]² becomes 1/2k(6k²+ 3k − 1) + 9(k+1)² - 6(k+1) + 1
Can anybody explain?

 

[tkhunny]

It's nice if you can get it in the same form, but it is unnecessary. Just prove the two expressions equivalent. Maybe expand both?

 

[JeffM]

We have a question which says:-

Prove for any natural number n that:-

2² + 5² + 8² + ··· + (3n − 1)² = 1/2n(6n² + 3n − 1)

and here is the solution:-
P(n): 2²+5²+8²+ . . .+ (3n-1)² = 1/2n(6n² + 3n − 1)

For 'n' = 1, P(1) = 2² = 1/2(1)[6(1)²+3(1) -1 ]
4 = 1/2(1)[6(1)²+3(1)-1]
4 = 1/2 * 8
4 = 4
LHS = RHS
So, P(1) is true.

Assume that P(k) is true for some integer 'k'
P(k):- 2²+5²+8²+ . . .+ (3k-1)² = 1/2k(6k² + 3k − 1)------------ eq(1)

We shall now prove that P(k+1) is true. Now we have:-
2²+5²+8²+ . . .+ (3k-1)² +[3(k+1)-1]² = 1/2k(6k² + 3k − 1) +[3(k+1) -1]²

using LHS,
=> 2²+5²+8²+ . . .+ (3k-1)² +[3(k+1)-1]²
=> 1/2k(6k² + 3k − 1) + [3(k+1)-1]² ---- Using eq(1)
=> 1/2k(6k²+ 3k − 1) + 9(k+1)² - 6(k+1) + 1

Now how 1/2k(6k² + 3k − 1) + [3(k+1)-1]² becomes 1/2k(6k²+ 3k − 1) + 9(k+1)² - 6(k+1) + 1
Can anybody explain?

\(\displaystyle \text {Let } u = k + 1 \implies \{3(k + 1) - 1\}^2 = (3u - 1)^2 = (3u - 1)(3u - 1) =\)

\(\displaystyle 3u(3u - 1) - 1(3u - 1) = 9u^2 - 3u - 3u + 1 = 9u^ 2 - 6u + 1 \implies\)

\(\displaystyle \{3(k + 1) - 1\}^2 = 9(k + 1)^2 - 6(k + 1) + 1.\)

Of course, that does not end the proof.

 

[manishpamnani169]

It's nice if you can get it in the same form, but it is unnecessary. Just prove the two expressions equivalent. Maybe expand both?

Yes it may be unnecessary but it would be nice to gain a better understanding of what's happening here!!
Btw thanks for replying!

 

[manishpamnani169]

\(\displaystyle \text {Let } u = k + 1 \implies \{3(k + 1) - 1\}^2 = (3u - 1)^2 = (3u - 1)(3u - 1) =\)

\(\displaystyle 3u(3u - 1) - 1(3u - 1) = 9u^2 - 3u - 3u + 1 = 9u^ 2 - 6u + 1 \implies\)

\(\displaystyle \{3(k + 1) - 1\}^2 = 9(k + 1)^2 - 6(k + 1) + 1.\)

Of course, that does not end the proof.

Thank you very very much for the help!! Of course, that doesn't end the proof but it will be a great help to me as i was stuck on this problem for two days to figure out a solution!!
Thanks for answering and replying!!