3 math questions (summation, mathematical induction)

In summary, the two math questions are: 1) Prove by mathematical induction that sum of n terms is 3/4 - (2n+3)/2. 2) Prove by another method that summation is true when n=2.
  • #1
misogynisticfeminist
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2 math questions (summation, mathematical induction)

I have 2 questions regarding summation and mathematical induction

2. Prove by mathematical induction
[tex] \sum^n_{r=1} \frac {1}{r(r+2)} = \frac {3}{4} - \frac {(2n+3)}{2(n+1)(n+2)} [/tex]
i am now trying to prove that

[tex] 3/4 - \frac{2(n+1)+3}{2(n+2)(n+3)} - \frac {1}{n+1}{n+3} [/tex]

I seem to get caught up with the algebra and what i get is

[tex] \frac {2n^2+5n+1}{2(n+1)(n+2)(n+3)} [/tex]. instead of [tex] \frac {2n^2+5n+3}{2(n+1)(n+2)(n+3)} [/tex]

3. Prove by induction that [tex] \sum^n_{r=1}rx^{r-1}=\frac{1-(n+1)x^n+nx^{n+1}}{(1-x)^2} [/tex]

again, I'm caught up with the algebra.
 
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  • #2
I use a different method:

1) proof summation is true when n=2
2) consider n=k+1.
[tex]\sum^{k+1}_{r=1} \frac{1}{r(r+2)} = \frac{3}{4} - \frac{2(k+1)+3}{2(k+2)(k+3)}[/tex]
[tex]= \sum^k_{r=1} \frac{1}{r(r+2)} + \mbox{(k+1)^{th}} \mbox{term}[/tex]

Well, it works. Sorry if I'm not very helpful :uhh:
 
  • #3
"i am now trying to prove that
[tex] 3/4 - \frac{2(n+1)+3}{2(n+2)(n+3)} - \frac {1}{n+1}{n+3} [/tex]"

This makes no sense- it is neither an equation nor a statement- it is a formula. What are you trying to prove about it?

What you should be trying to prove is that
[tex]\frac {3}{4} - \frac {(2n+3)}{2(n+1)(n+2)}+ \frac{1}{(n+1)(n+3)} = \frac{3}{4}- \frac{(2n+5)}{2(n+2)(n+3)}[/tex]
The left side of that is just the formula for the sum to n plus the n+1 term and the right side is the same formula for the sum to n+1.
Now to show they are the same: obviously the "3/4" cancel. Get the common denominator for the two fractions left on the left side:
[tex]-\frac {(2n+3)(n+3)}{2(n+1)(n+2)(n+3)}+ \frac{2(n+2)}{2(n+1)(n+2)(n+3)}[/tex]
[tex]-\frac{(2n^2+ 9n+ 9)+(2n+4)}{2(n+1)(n+2)(n+3)}[/tex]
[tex]-\frac{2n^2+7n+5}{2(n+1)(n+2)(n+3)}[/tex]
In order to have the same denominator as on the right, we obviously need to cancel the "n+1" in the denominator and, sure enough, 2n2+ 7n+ 5= (n+1)(2n+ 5)!

As far as 2 is concerned, are you required to use induction? I notice that the left side is just the derivative of 1+ x+ x2+...+ xn and I know that xn+1- 1= (x- 1)(1+ x+ x2+...+ xn) so that
[tex]1+ x+ x^2+...+ x^n= \frac{x^n-1}{x-1}[/tex]. You can get the result by using the quotient rule to differentiate the right side.

Of course, if you MUST us induction, then you need to show that
[tex]\frac{1-(n+1)x^n+nx^{n+1}}{(1-x)^2}+ (n+1)x^n= \frac{1-(n+2)x^2+(n+1)x^{n+2}}{(1-x)^2} [/tex]

Again, concetrate on the right hand side. You will need to multiply that "(n+1)xn" term by (1-x)2/(1-x)2 and then combine the numerators.
 

1. What is summation in math?

Summation, also known as sigma notation, is a way to represent the sum of a series of numbers or terms. It uses the symbol Σ, which stands for "sum", followed by the starting value of the series and the ending value, with the expression to be summed written to the right of the symbol.

2. How do you solve summation problems?

To solve a summation problem, you can use the formula Σn = (n(n+1))/2, where n represents the number of terms in the series. You can also use algebraic techniques, such as factoring and simplifying, to solve more complex summation problems.

3. What is mathematical induction?

Mathematical induction is a method of proving mathematical statements or theorems that are true for an infinite number of cases. It involves two steps: the base case, where the statement is proven to be true for a specific value, and the inductive step, where the statement is proven to be true for the next value assuming it is true for the previous value.

4. When should mathematical induction be used?

Mathematical induction should be used when trying to prove a statement or theorem that applies to an infinite number of cases, such as with summation problems. It is also a useful tool in proving theorems in number theory and combinatorics.

5. What is the difference between strong and weak mathematical induction?

The difference between strong and weak mathematical induction lies in the inductive step. In weak induction, the statement is proven to be true for the next value assuming it is true for only the previous value. In strong induction, the statement is proven to be true for the next value assuming it is true for all previous values. Strong induction is often used when the statement builds upon itself, while weak induction is used for simpler statements.

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