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- Feb 26th 2007, 11:49 AMchillerbros17Proofs by Mathematical Induction
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- Feb 26th 2007, 03:37 PMtopsquark
Try this for n = 1:

1*3 =? 1(1 + 1)(2*1 +7)/6

3 =? 1*2*9/6

3 =? 3 (Check!)

Now assume the statement is true for some integer k. That is

1*3 + 2*4 + ... + k(k + 2) = k(k + 1)(2k + 7)/6

We need to show that it also holds for k + 1:

1*3 + 2*4 + ... + k(k + 2) + (k + 1)(k + 3) = (k + 1)(k + 2)(2k + 9)/6

We know that 1*3 + ... + k(k + 2) = k(k + 1)(2k + 7)/6 so use that in the above equation. Thus

k(k + 1)(2k + 7)/6 + (k + 1)(k + 3) = (k + 1)(k + 2)(2k + 9)/6

should be an identity.

Multiply both sides by 6, then simplify:

k(k + 1)(2k + 7) + 6(k + 1)(k + 3) = (k + 1)(k + 2)(2k + 9)

(k^2 + k)(2k + 7) + 6(k^2 + 4k + 3) = (k^2 + 3k + 2)(2k + 9)

2k^3 + 9k^2 + 7k + 6k^2 + 24k + 18 = 2k^3 + 15k^2 + 31k + 18

2k^2 + 15k^2 + 31k + 18 = 2k^3 + 15k^2 + 31k + 18

0 = 0 (Check!)

Thus the proposition is true for k = 1, so it is true for all integer k >= 1.

-Dan - Feb 26th 2007, 08:39 PMCaptainBlack
Let P(n) be the proposition: 57 divides 7^(n+2) + 8^(2n+1).

(a) P(1) is true as 7^3+8^3 = 855 = 15*57

(b) Now suppose for some k>=1, P(k) is true. Then 7^(k+2) + 8^(2k+1) = a*57

for some positive integer a, or 8^(2k+1) = a*57 - 7^(k+2).

Now:

7^((k+1)+2) + 8^(2(k+1)+1) = 7*7^(k+2) + 64*8^(2k+1)

........................................= 7*7^(k+2) + 64*[a*57 - 7^(k+2)]

........................................= [7 - 64]* 7^(k+2) + 64*a*57

........................................= 57*[ -7^(k+2) + 64*a]

Hence 7^((k+1)+2) + 8^(2(k+1)+1) is divisible by 57, which proves P(k+1),

therfore with (a) this proves P(n) is true for all n>=1 by mathematical

induction.

RonL - Feb 27th 2007, 10:24 AMSoroban
Hello, chillerbros17!

Another approach to #1 . . .

Quote:

1) Prove by Mathematical Induction:

. . 1·3 + 2·4 + 3·5 + . . . + n(n + 2) .= .n(n + 1)(2n + 7)/6 .for all integers__>__1

Verify S(1): .1·3 .= .1(2)(9)/6 . → . 3 = 3 . . . true!

Assume S(k) is true: .1·3 + 2·4 + 3·5 + ... + k(k + 2) .= .k(k + 1)(2k + 7)/6

We want to prove that S(k+1) is true:

. . 1·3 + 2·4 + 3·5 + ... + (k + 1)(k + 3) .= .(k + 1)(k + 2)(2[k+1] + 7)/6 .**[1]**

written out here*for reference*... so we'll recognize it later.

Begin with S(k): .1·2 + 2·4 + 3·5 + ... + k(k + 2) .= .k(k + 1)(2k + 7)/6

Add (k + 1)(k + 3) to both sides:

. . 1·3 + 2·4 + 3·5 + ... + (k + 1)(k + 3) .= .k(k + 1)(2k + 7)/6 + (k + 1)(k + 3)

The left side is the left side of [1].

We must show that the right side is the right side of [1].

. . . . . . . . . . . . . k(k + 1)(2k + 7) . . . . . . . . . . . . . . . k(k + 1)(2k + 7) + 6(k + 1)(k + 3)

The right side is: .-------------------- + (k + 1)(k + 3) .= . -----------------------------------------

. . . . . . . . . . . . . . . . . . 6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

. . . . . . .(k + 1) [k(2k + 7) + 6(k + 3)] . . . (k + 1)(2kČ + 13k + 18)

Factor: . ------------------------------------ .= .-----------------------------

. . . . . . . . . . . . . . . . . 6 . . . . . . . . . . . . . . . . . . . 6

. . . . . (k + 1)(k + 2)(2k + 9) . . .(k + 1)(k + 2)(2[k+1] + 7)

. . . = .-------------------------- .= .--------------------------------

. . . . . . . . . . . . 6 . . . . . . . . . . . . . . . . . . 6

This is the right side of [1] . . . The induction proof is complete.

- Mar 3rd 2007, 08:42 AMPriyasum of 1^k+ 2^k+ 3^k+ 4^k+...n^k
could anyone help me out for this question...

i think we need to use the general terms for the sum of n terms, sum of 1^2+ 2^2+...n^2..etc - Mar 3rd 2007, 08:52 AMtopsquark
- Mar 3rd 2007, 11:29 PMCaptainBlack
A complicated topic see the Power Sum article on MathWorld for details.

RonL