# Discrete Math

Use mathematical induction to establish the following formula.

n
Σ i² / [(2i-1)(2i+1)] = n(n+1) / 2(2n+1)
i=1

Thanks for any helpful replies :)

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1. check if true for n=1
LS = 1/(1(3)) = 1/3
RS = 1(2)/(2(3) = 2/6 = 1/3 , checks!

Assume it to be true for n = k
that is ....
1/(1(3)) + 4/(3(5)) = .... k^2/((2k-1)(2k+1)) = k(k+1)/(2(2k+1))

prove then it must be true for n = k+1
that is
1/(1(3)) + 4/(3(5)) + .. + (k+1)^2/((2k+1)(2k+3)) = (k+1)(k+2)/(2(2k+3))

LS = k(k+1)/2(2k+1)) + (k+1)^2/((2k+1)(2k+3))

=[k(k+1)(2k+3) + 2(k+1)^2]/[2(2k+1)(2k+3)]
= (k+1)[k)2k+3) + 2(k+1)]/[2(2k+1)(2k+3)]
= (k+1)[2k^2 + 5k + 2]/[2(2k+1)(2k+3)]
= (k+1)(2k+1)(k+2)/[2(2k+1)(2k+3)]
= (k+1)(k+2)/(22k+3)
= RS

QED!

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2. 6th line should have been

1/(1(3)) + 4/(3(5)) + .... + k^2/((2k-1)(2k+1)) = k(k+1)/(2(2k+1))

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3. Ok thank you for your helpful response! I have a couple of questions though. . .

Is the 15th line suppose to be '(k+1)(k+2)/(22k+3)'?

Also, the 16th line = RS, which is what exactly?

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4. line 15th, clearly a typo, was hoping you would realize it was
= (k+1)(k+2)/(2(2k+3))

in 1/(1(3)) + 4/(3(5)) + .. + (k+1)^2/((2k+1)(2k+3)) = (k+1)(k+2)/(2(2k+3))
there is a left side (LS) and a right side (RS) of the equation.
I started with LS and proved that it equals the RS, thus showing that the equation is true.

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5. Yea that's what I thought. . .Hey if you don't mind helping me further I have been working on this problem for a while and I am a bit stuck. IDK where to go from here or if I am doing it correctly:

Use mathematical induction to prove the truth of each of the following assertions for all n ≥1.

5^2n – 2^5n is divisible by 7

If n = 1, then 5^2(1) - 2^5(1) = -7, which is divisible by 7. For the inductive case, assume k ≥ 1, and the result is true for n = k; that is 7 | (5^2k + 2^5k). Use the assumption to prove n = k + 1, in other words, 5^(2(k + 1)) - 2^(5(k + 1)) is divisible by 7. Now,

5^(2(k + 1)) - 2^(5(k + 1))
= 5^(2k + 2) - 2(5k + 5)
= 5^(2k) · 5^2 - 2^(5k) · 2^5
= 25 · 5^(2k) - 32 · 2^(5k)
= IDK what to do from here. . .

Any suggestions? Thank you again!

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6. Any suggestions?

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