By using induction, verify that each equation below is true for very positive integer n.

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By using induction, verify that each equation below is true for very positive integer n.

[From: ] [author: ] [Date: 12-12-08] [Hit: ]

!!!So when you neglect to check step 1, you can proove that:For n=0, 2n+1=1 →→→ 1 is even!......

Note: Pi R Square is incorrect because of missing first step.

Example of proof of the impportance of first step:
Let's prove that, for any integer n,
2n+1 is even

Second step: Let's suppose the property is true for rank n:
2n+1 is even

Third step: Let's prove the property is true for rank n+1:
2(n + 1) + 1 = 2n + 2 + 1
= 2n + 1 + 2
As 2n+1 is even (because of step 2), then must (2n+1)+2 be even
(since adding 2 to any even integer will yield an even integer)
Thus 2n+3=2(n+1)+1 is even
And the property is proven!!!!!

So when you neglect to check step 1, you can proove that:
For n=0, 2n+1=1 →→→ 1 is even!!!!!
For n=1, 2n+1=3 →→→ 3 is even!!!!!
For n=2, 2n+1=5 →→→ 5 is even!!!!!

So Pi R Squared solving is incorrect.

-

Hi,

Assume that 1^2 + 2^2 +3^2+ ....+ n^2 = (n(n+1)(2n+1)) / 6 is true.

Then the next step would be:

1^2 + 2^2 +3^2+ ....+ n^2 + (n + 1)² = (n(n+1)(2n+1)) / 6 + (n + 1)²

(n(n+1)(2n+1)) / 6 + (n + 1)² =

(n(n+1)(2n+1)) / 6 + n² + 2n + 1 =

(n(n+1)(2n+1)) / 6 + 6(n² + 2n + 1)/6 =

(n(n+1)(2n+1)) + 6(n² + 2n + 1)
--------------------------------------… =
. . . . . . . . . . . 6

(n(2n² + 3n + 1)) + 6n² + 12n + 6
--------------------------------------… =
. . . . . . . . . . . 6

2n³ + 3n² + n + 6n² + 12n + 6
--------------------------------------… =
. . . . . . . . . 6

2n³ + 9n² + 13n + 6
----------------------------- =
. . . . . . . 6

2n³ + 9n² + 13n + 6 is divisible by n + 1
. _____________
-1)2 + 9 + 13 + 6
. ___-2___-7_ -6_
. .2 . 7 . . .6 . . 0

2n³ + 9n² + 13n + 6 = (n + 1)(2n² + 7n + 6)

But 2n² + 7n + 6 is also divisible by n + 2
. _________
-2)2 + 7 + 6
. ___-4__-6_
. .2 . 3 . . 0

2n² + 7n + 6 = (n + 2)(2n + 3)

So,
2n³ + 9n² + 13n + 6 = (n + 1)(n + 2)(2n + 3), which is the same as:

2n³ + 9n² + 13n + 6 = (n + 1)(n + 2)(2(n+1) + 1)

So 1^2 + 2^2 +3^2+ ....+ n^2 + (n + 1)² = (n + 1)(n + 2)(2(n+1) + 1) works for n + 1 terms, just as

1^2 + 2^2 +3^2+ ....+ n^2 = (n(n+1)(2n+1)) / 6 worked for n terms.

I hope that helps!! :-)

12

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