Thread: Some help on mathematical induction?

1. Some help on mathematical induction?

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

2. 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.

3. 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?

4. Originally Posted by chiph588@
What is $r$ here?
Sigma Notation: n (on top) r=1 (on bottom)

5. 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.

6. 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.

7. Originally Posted by blaze
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?

8. 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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1/3 1/15 1/35 to nterms find the sum

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