1. ## mathematical induction

Hi!!! can anyone help me with these mathematical inductions? normally, inductions are given to prove. but these are different. need HELP!!!

2. ## Re: mathematical induction

Originally Posted by jkjk

Hi!!! can anyone help me with these mathematical inductions? normally, inductions are given to prove. but these are different. need HELP!!!
I think the idea is to try to find the formulas yourself. For instance, the first few examples would be
$\displaystyle \sum_{k = 1}^{2} (2k - 1) = (2(1) -1) + (2(2) - 1) = 1 + 3 = 4$

$\displaystyle \sum_{k = 2}^{4} (2k - 1) = (2(2) -1) + (2(3) - 1) + (2(4) - 1) = 3 + 5 + 7 = 15$

$\displaystyle \sum_{k = 3}^{6} (2k - 1) = (2(3) -1) + (2(4) - 1) + (2(5) - 1) + (2(6) - 1)$ $\displaystyle = 5 + 7 + 9 + 11 = 32$

You may not be able to see the series from this but I'll give you a hint: $\displaystyle S(n) = an^2 + bn + c$. You've got three equations and three unknowns.

-Dan

3. ## Re: mathematical induction

the answer is 3n^2 +2n-1 .. but i don't get it

4. ## Re: mathematical induction

It always help to look at different small values for n.

$\displaystyle \sum_{k=1}^{2*1}(2k - 1) = \sum_{k=1}^2(2k - 1) =$

$\{(2 * 1) - 1\} + \{(2 * 2) - 1\} = 1 + 3 = 4 = 2^2 = (2 * 1)^2 - 0.$

$\displaystyle \sum_{k=2}^{2*2}2k - 1) = \sum_{k=2}^4(2k - 1) =$

$\{(2 * 2) - 1\} + \{(2 * 3) - 1\} + \{(2 * 4) - 1\} = 3 + 5 +7 = 15 = 16 - 1 = (2 * 2)^2 - 1.$

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

$\{(2 * 3) - 1\} + \{(2 * 4) - 1\} + \{(2 * 5) - 1\} +\{(2 * 6) - 1\} = 5 + 7 + 9 + 11 = 32 = (2 * 3)^2 - 4.$

Does this remind you of anything?

I'll give you another hint.

$\displaystyle \sum_{i=1}^m(2i - 1) = what?$

5. ## Re: mathematical induction

(2n(2n+1)(4n+1))/6 -(n+1)

6. ## Re: mathematical induction

Originally Posted by jkjk
(2n(2n+1)(4n+1))/6 -(n+1)
Both topsquark and I calculated that if n = 3, the sum is 32.

$\dfrac{2 * 3(2 *3 + 1)(4 * 3 + 1)}{6} - (3 + 1) = 7 * 13 - 4 = 87 \ne 32.$

7. ## Re: mathematical induction

then i am not clever enough to do that

8. ## Re: mathematical induction

Originally Posted by jkjk
then i am not clever enough to do that
Not clever enough to do what? He and I just did some arithmetic first. Nothing clever about that.

Then topsquark, who is cleverer than I, saw that a quadratic would do the trick without any apparent intermediate steps. But I saw that this sum is related to the sum of the first p odd numbers and so started looking for squares too.

He gave you an analytic approach. I gave you a pattern recognition approach. What is it that you are not seeing?

9. ## Re: mathematical induction

$\displaystyle \sum_{k = n}^{2n} (2 k - 1) = \text{?}$.

Define $\displaystyle f(m) = \sum_{k = 1}^{m} (2 k - 1)$.

Then $\displaystyle \sum_{k = n}^{2n} (2 k - 1) = f(2n) - f(n-1)$, because always have (for $\displaystyle n \ge 2$) that..... $\displaystyle \sum_{k = 1}^{n-1} a_k + \sum_{k = n}^{2n} a_k = \sum_{k = 1}^{2n} a_k$.

So to find $\displaystyle \sum_{k = n}^{2n} (2 k - 1)$, it suffices to find $\displaystyle f(m)$.

But $\displaystyle f(m) = \sum_{k = 1}^{m} (2 k - 1) = \sum_{k = 1}^{m} (2 k) + \sum_{k = 1}^{m} (- 1) = 2 \sum_{k = 1}^{m} k - \sum_{k = 1}^{m} (1)$.

Thus $\displaystyle f(m) = 2 \left( \sum_{k = 1}^{m} k \right) - m$.

I bet you've seen $\displaystyle \sum_{k = 1}^{m} k$ before, and if so, you can find $\displaystyle f(m)$, and so you can get a formula for $\displaystyle \sum_{k = n}^{2n} (2 k - 1) = f(2n) - f(n-1)$.

10. ## Re: mathematical induction

I would work from the sum of the first $n$ natural numbers, $\sum_{k=1}^n k$, which, as johnsomeone says, you ought to have seen before. Given that, you have $$\sum_{k=1}^n (2k - 1) =\sum_{k=1}^{2n} k - \sum_{k=1}^n 2k$$ which just says that the sum of the first $n$ odd numbers is the sum of the first $2n$ natural numbers less the sum of the first $n$ even numbers.

Having said that, if you work out the first few terms $\sum_{k=1}^n (2k - 1)$, you should very definitely see a pattern.

11. ## Re: mathematical induction

Hi jkjk,
Frequently, formal manipulations with sums are helpful; what you might call "Sigma algebra".

12. ## Re: mathematical induction

Originally Posted by jkjk

Hi!!! can anyone help me with these mathematical inductions? normally, inductions are given to prove. but these are different. need HELP!!!
You should know that the sum of the first N odd integers is N^2, so

\displaystyle \begin{align*} \sum_{k = 1}^{N}{ \left( 2k - 1 \right) } &= N^2 \\ \\ \sum_{k = 1}^{2n} \left( 2k - 1 \right) &= \left( 2n \right) ^2 \\ &= 4n^2 \\ \\ \sum_{k = 1}^{n - 1} \left( 2k - 1 \right) &= \left( n - 1 \right) ^2 \\ &= n^2 - 2n + 1 \\ \\ \sum_{ k = n}^{2n} \left( 2k - 1 \right) &= \sum_{ k = 1 }^{2n} \left( 2k - 1 \right) - \sum_{k= 1}^{n - 1} \left( 2k - 1 \right) \\ &= 4n^2- \left( n^2 - 2n + 1 \right) \\ &= 3n^2 + 2n - 1 \end{align*}

Now you could try induction if you want...

13. ## Re: mathematical induction

THANKS everyone..