Pre-Calculus, anybody????? I need help!!!!!?

Question

Show that the sum of the first n positive odd integers is n^2.

Can you show all the steps to the process and show all of the steps? The most complete and detailed answer will receive 10 points. Thank you.

Answer 1

Let's see.

Suppose the biggest odd number is m. The numbers are 1, 3, 5, .... m.
The first number in the series is 1. The second is 1 + 2. The third is is 1 + 2*2. The last one, m, is 1 + (n-1)*2. I hope this is obvious.

Now

Write them in ascending order:

1 + 3 + 5 + ... + m

Now write them in descending order:

m + (m-2) + (m-4) + ... + 1

Add these together:

(m+1) + (m+1) + (m+1) + .... + (m+1)

We have n of these terms. So this sum is n(m+1). We'll have to divide by 2 to get the number we're really looking for, because we included every term twice.

So the answer is (n/2)(m+1).

But m = 1+(n-1)*2, so the sum is:

(n/2)(1 +(n-1)*2 + 1) = (n/2)(2 + 2n - 2) = (n/2)(2n) = n²

Pretty neat!

Answer 2

Can be done by the fact that the sum of it's terms is equal to its arithmetic mean.
1+3+5+7+9+11+...+(2n-1) = n[(1+(2n-1))/2] = n^2

Answer 3

1=1
1+3=2^2
let it be true for some n=k
1+3+.... k odd nos=k^2
1+3+.... k+1 odd nos==k^2+1+(k+1-1)*2
=k^2+2k+1
=(k+1)^2
so it is true for any n n being element of N

Answer 4

Prove by the principle of mathematical induction.

Use the same method as you would to prove that the sum of integers 1 through n = n(n+1)/2

http://en.wikipedia.org/wiki/Mathematical_induction#Example






.

Answer 5

word: it incredibly is a trouble-free mistake to think of that the fourth means of a sum is the sum of the fourth powers. (radical(x) + 8)^4 isn't comparable to (radical(x))^4 + 8^4. f(g(h(x))) = f(g(radical(x))) = f(radical(x) + 8) = (radical(x) + 8)^4 + 2 = (4 p.c. 0)(radical(x))^4 + (4 p.c. a million)(8)(radical(x))^3 + (4 p.c. 2)(8^2)(radical(x))^2 + (4 p.c. 3)(8^3)(radical(x)) + (4 p.c. 4)(8^4) + 2 = a million(x^2) + 4(8)x*radical(x) + 6(sixty 4)x + 4(512)(radical(x)) + a million(4096) + 2 = x^2 + 32x*radical(x) + 384x + 2048(radical(x)) + 4098 Lord bless you immediately!

Answer 6

Sum (2n-1) for n=1 to infinity.