Some help on mathematical induction?

**blaze** · November 29th 2009, 09:06 AM

Have some trouble completing these questions:
Any help will be greatly appreciated:

1 Use mathematical induction to prove that 5^n+9^n+2 is divisible by 4 for all n>0

2a) use mathematical induction to prove that:
1/(2r-1)(2r+1)=n/2n+1 for all n>0
b) Hence show that the sum of the first (n +1) terms of the series
1/3+1/15+1/35+1/63+....is (n+1)/(2n+3)

3) (1)(1!)+(2)(2!)+(3)(3!)+...+(n)(n!)=(n+1… where n>0 Find the minimum number of terms of the series for the sum to exceed 10^9

**chiph588@** · November 29th 2009, 12:03 PM

1.) Base case of $n=0$ : $5^0+9^0+2 = 4$ .

Now suppose $5^n+9^n+2$ is divisible by $4$ , then $5^n+9^n+2 = 4k$ for some natural number $k$ .

Then $4 \cdot 5^n+8 \cdot 9^n+5^n+9^n+2 = 4 \cdot 5^n+8 \cdot 9^n+4k$ .

Upon simplification we get $5^{n+1}+9^{n+1}+2 = 4(k+5^n+2\cdot9^n) = 4k'$ . And we're done.

**chiph588@** · November 29th 2009, 12:05 PM

Originally Posted by blaze

2a) use mathematical induction to prove that:
1/(2r-1)(2r+1)=n/2n+1 for all n>0
b) Hence show that the sum of the first (n +1) terms of the series
1/3+1/15+1/35+1/63+....is (n+1)/(2n+3)

What is $r$ here?

**blaze** · November 29th 2009, 10:04 PM

Originally Posted by chiph588@

What is $r$ here?

Sigma Notation: n (on top) r=1 (on bottom)

**Prove It** · November 30th 2009, 01:03 AM

Originally Posted by blaze

Have some trouble completing these questions:
Any help will be greatly appreciated:

1 Use mathematical induction to prove that 5^n+9^n+2 is divisible by 4 for all n>0

2a) use mathematical induction to prove that:
1/(2r-1)(2r+1)=n/2n+1 for all n>0
b) Hence show that the sum of the first (n +1) terms of the series
1/3+1/15+1/35+1/63+....is (n+1)/(2n+3)

3) (1)(1!)+(2)(2!)+(3)(3!)+...+(n)(n!)=(n+1… where n>0 Find the minimum number of terms of the series for the sum to exceed 10^9

So for 2) you're trying to prove

$\sum_{r = 1}^n\frac{1}{(2r - 1)(2r + 1)} = \frac{n}{2n + 1}$ for all $n > 0$ ?

Base step: $n = 1$ .

$LHS = \sum_{r = 1}^n\frac{1}{(2r - 1)(2r + 1)}$

$= \frac{1}{3}$ .

$RHS = \frac{1}{3}$

$= LHS$ as required.

Inductive step: Assume the statement is true for $n = k$ .

So $\sum_{r = 1}^k\frac{1}{(2r - 1)(2r + 1)} = \frac{k}{2k + 1}$ .

We need to show that the equation holds true for $n = k + 1$ .

$LHS = \sum_{r = 1}^{k + 1}\frac{1}{(2r - 1)(2r + 1)}$

$= \sum_{r = 1}^k\frac{1}{(2r - 1)(2r + 1)} + \frac{1}{(2k + 1)(2k + 3)}$

$= \frac{k}{2k + 1} + \frac{1}{(2k + 1)(2k + 3)}$

$= \frac{k(2k + 3)}{(2k + 1)(2k + 3)} + \frac{1}{(2k + 1)(2k + 3)}$

$= \frac{2k^2 + 3k + 1}{(2k + 1)(2k + 3)}$

$= \frac{(2k + 1)(k + 1)}{(2k + 1)(2k + 3)}$

$= \frac{k + 1}{2k + 3}$ .

$RHS= \frac{k + 1}{2k + 3}$

$= LHS$ as required.

Q.E.D.

**blaze** · November 30th 2009, 01:47 AM

Originally Posted by Prove It

So for 2) you're trying to prove

$\sum_{r = 1}^n\frac{1}{(2r - 1)(2r + 1)} = \frac{n}{2n + 1}$ for all $n > 0$ ?

Base step: $n = 1$ .

$LHS = \sum_{r = 1}^n\frac{1}{(2r - 1)(2r + 1)}$

$= \frac{1}{3}$ .

$RHS = \frac{1}{3}$

$= LHS$ as required.

Inductive step: Assume the statement is true for $n = k$ .

So $\sum_{r = 1}^k\frac{1}{(2r - 1)(2r + 1)} = \frac{k}{2k + 1}$ .

We need to show that the equation holds true for $n = k + 1$ .

$LHS = \sum_{r = 1}^{k + 1}\frac{1}{(2r - 1)(2r + 1)}$

$= \sum_{r = 1}^k\frac{1}{(2r - 1)(2r + 1)} + \frac{1}{(2k + 1)(2k + 3)}$

$= \frac{k}{2k + 1} + \frac{1}{(2k + 1)(2k + 3)}$

$= \frac{k(2k + 3)}{(2k + 1)(2k + 3)} + \frac{1}{(2k + 1)(2k + 3)}$

$= \frac{2k^2 + 3k + 1}{(2k + 1)(2k + 3)}$

$= \frac{(2k + 1)(k + 1)}{(2k + 1)(2k + 3)}$

$= \frac{k + 1}{2k + 3}$ .

$RHS= \frac{k + 1}{2k + 3}$

$= LHS$ as required.

Q.E.D.

I already got the answer down- I need 2b) and 3) please

**Jameson** · November 30th 2009, 01:50 AM

Originally Posted by blaze

I already got the answer down- I need 2b) and 3) please

If you understand 2a then there is no way you didn't do 2b along the way. The proof of 2a by induction requires showing 2b. Please post one question per thread in the future and show what work you have attempted.

So what do you need help with for 2?

**blaze** · November 30th 2009, 02:34 AM

I don't get how you would pose the answer for 2b)

As for 3), I have proved the mathematical induction theory but not sure how to " Find the minimum number of terms of the series for the sum to exceed 10^9"

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