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I am doing mathematical induction. I am stuck with the question below. The left hand side is not getting equal to the right hand side.
Please guide me how to do it further.

$1^2 + 3^2+ 5^2 + \cdots + (2n-1)^2 = \frac{1}{3}n(2n-1)(2n+1)$.

Sol:

$P(n):\ 1^2 + 3^2 + 5^2 + \cdots + (2n-1)^2 = \frac{1}{3}n(2n-1)(2n+1)$.

For $n=n_1 = 1$

$$P(1) = \frac{1}{3}{3} = (1)^2.$$ Hence it is true for $n=n_0 = 1$.

Let it be true for $n=k$ $$P(k): 1^2 + 3^2 + 5^2 + \cdots + (2k-1)^2 = \frac{1}{3}k(2k-1)(2k+1).$$ We have to prove that it is true for $P(k+1)$. $$P(k+1) = 1^1+3^2+5^2+\cdots+(2k+1)^2 = \frac{1}{3}(k+1)(2k+1)(2k+3)\tag{A}.$$

Taking LHS: $$\begin{align*} 1^2 + 3^2 + 5^2 + \cdots + (2k+1)^2 &= 1^2+3^2+5^2+\cdots + (2k+1)^2\\ &= 1^2 + 3^2 + 5^2 + \cdots + (2k-1)^2 + (2k+1)^2\\ &= \frac{1}{3}k(2k-1)(2k+1) + (2k+1)^2\\ &=\frac{k(2k-1)(2k+1)+3(2k+1)^2}{3}\\ &=\frac{(2k+1)}{3}\left[k(2k-1) + 3(2k+1)\right]\\ &=\frac{(2k+1)}{3}\left[2k^2 - k + 6k + 3\right]\\ &=\frac{1}{3}(2k+1)(2k^2 +5k + 3)\\ &=\frac{1}{3}(2k+1)(k+1)\left(k+\frac{3}{2}\right) \tag{B} \end{align*}$$

EDIT:

Solving EQ (A):

$=(1/3)(2k^2+5K+3) (2K+1) \tag{C}$

Comparing EQ(B) and EQ(C)

Hence proved that it is true for $n = k+1.$

Thus the proposition is true for all $n >= 1$.

Thanks.

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@Arturo: Your answer is more detailed than mine. In a case like this, don't you agree one should write an answer instead of a comment since there's not much more to say and the question is likely to remain unanswered otherwise? –  joriki Oct 14 '11 at 16:58
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@joriki: I was kind of hoping the OP might answer his own question once he got the right answer... –  Arturo Magidin Oct 14 '11 at 17:02
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@Arturo: a) Wow, typesetting all that stuff shows true dedication :-). b) When I said your answer is more detailed than mine, I meant that it's more helpful -- why did you remove it? –  joriki Oct 14 '11 at 17:14
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I'm with Arturo: since we're allowing people to answer their own questions anyway, it seems to be a good thing to coax people into answering their own questions with appropriate nudges... –  J. M. Oct 14 '11 at 17:16
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@J. M.: It seems to me that this is based on an unrealistic optimism about how people deal with this site. I see several questions every day that don't get answered because someone wrote the answer in a comment, and I think I have yet to see even one question where the OP answered their own question because someone provided the answer in a comment. –  joriki Oct 14 '11 at 17:21
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2 Answers 2

up vote 4 down vote accepted

Everything is OK except for the very last line. You somehow lost a factor of two. The penultimate line is already the result you want, since $2k^2+5k+3=(k+1)(2k+3)$.

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You beat me to it! I think I know how the factor of 2 got `lost': the OP solved the quadratic equation $2k^2+5k+3=0$, giving the two roots $k=-3/2$ and $k=-1$, which suggests the (incorrect) factorisation $(k+3/2)(k+1)$. The moral of the story is that there is a difference between factorising an expression and solving an equation, even if the two processes are intimately related. –  Shane O Rourke Oct 14 '11 at 17:08
    
@joriki: Thanks, I actually used a calculator to solve that and as Shane mentioned it gave the the above results.. –  Fahad Uddin Oct 14 '11 at 17:11
    
@ShaneORourke: Thanks a lot. I had mostly been solving quadratic equation instead of factorizing them. I think this is the first time it gave different result. –  Fahad Uddin Oct 14 '11 at 17:13
    
@Akito: I'm not sure I understand your first comment. You used a calculator to solve what? The quadratic equation that Shane wrote? And it gave which above results? The two roots that Shane wrote? Your statement "it gave the above results" seems to imply that you think there's something about those results, but as Shane explained, it's not those results that are wrong but how you interpreted them and what you did with them. –  joriki Oct 14 '11 at 17:16
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@Akito: If $r_1$ and $r_2$ are the roots of $ax^2+bx+c$, then $(x-r_1)(x-r_2)$ can only equal $ax^2+bx+c$ if $a=1$ (just look at the leading coefficient!). In general, $$(x-r_1)(x-r_2) = x^2 + \frac{b}{a}x + \frac{c}{a}.$$Every time your quadratic is not monic, the method you used will fail; you have to keep track of that leading coefficient. –  Arturo Magidin Oct 14 '11 at 17:18
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NOTE: I am not saying anything different, before someone start commenting that my answer is not any different.

$$ \begin{align*} 1^2 + 3^2 + 5^2 + \cdots + (2k+1)^2 &= 1^2+3^2+5^2+\cdots + (2k+1)^2\\ &= 1^2 + 3^2 + 5^2 + \cdots + (2k-1)^2 + (2k+1)^2\\ &= \frac{1}{3}k(2k-1)(2k+1) + (2k+1)^2\\ &=\frac{k(2k-1)(2k+1)+3(2k+1)^2}{3}\\ &=\frac{(2k+1)}{3}\left[k(2k-1) + 3(2k+1)\right]\\ &=\frac{(2k+1)}{3}\left[2k^2 - k + 6k + 3\right]\\ &=\frac{1}{3}(2k+1)(2k^2 +5k + 3)\\ &=\frac{1}{3}(2k+1) \hspace{3pt}\left[(k+1)(2k+3)\right] \\ &= \frac{1}{3} (k+1)(2(k+1)-1)(2(k+1)+1) \end{align*} $$

The last line shows that the result is true for $n=k+1$

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Comment: this answer is not any different. –  Did Mar 8 '12 at 16:03
    
So I guess you did not read my NOTE: I am not saying anything different, before someone start commenting that my answer is not any different. –  Kirthi Raman Mar 10 '12 at 16:59
    
I did. $ $ $ $ $ $ –  Did Mar 10 '12 at 20:00
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