Proof with Induction 3/2-5/6+7/12-9/20+11/30-....

In summary, the conversation is about finding the general expression for a given sum and proving it by induction. The general expression is found using Newton's divided difference formula and is given by $\dfrac{(-1)^{n+1}(2n+1)}{n^2+n}$. The conversation also discusses the process of proving the expression using induction.
  • #1
Yankel
395
0
Hello all,

In the attached picture there is an equation. I need to fill the general expression on the left hand side, and to prove by induction that the sum is equal to the expression in the right hand side.

I am not sure how to find the general expression. Can you kindly assist ?

Thank you !

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  • #2
$\dfrac{(-1)^{n+1}(2n+1)}{n^2+n}$
 
  • #3
The numerators are clearly the odd numbers so: 2n+ 1.

The denominators are a little harder! I would have used "Newton's "divided difference" formula: adding a first term of "0", the "first differences" are 2- 0= 2, 6- 2= 4, 12- 6= 6, 20- 12= 8, 30- 20= 10; the "second differences" are 4- 2= 2, 6- 4= 2, 8- 6= 2, 10- 8= 2. Those are all "2" so all further "differences" are 0. The denominators are given by the quadratic [tex]0+ 2n+ (2/2)n(n-1)= n^2+ n[/tex].

Of course, since the +/- sign alternates we need -1 to a power. The first term, with n= 1, is positive so that can be either [tex](-1)^{n+1}[/tex] or [tex](-1)^{n-1}[/tex].
 
  • #4
I saw sequence of denominators, $2 ,6,12,20,30,...$, as

$(1\cdot 2), (2 \cdot 3), (3 \cdot 4), ( 4 \cdot 5),(5 \cdot 6), ... , [n \cdot (n+1)] , ...$
 
  • #5
I tried proving this by induction using the general statement that skeeter wrote, but I couldn't do it.

I am stuck at the n=k+1 stage...
 
Last edited:
  • #6
note $1 + \dfrac{(-1)^{n+1}}{n+1} = \dfrac{(n+1) + (-1)^{n+1}}{n+1}$
${\color{red}{\dfrac{3}{2} - \dfrac{5}{6} + \dfrac{7}{12} - \dfrac{9}{20} + ... + \dfrac{(-1)^{n+1}(2n+1)}{n(n+1)}}} + \dfrac{(-1)^{(n+1)+1}[2(n+1)+1]}{(n+1)[(n+1)+1]}$

${\color{red}\dfrac{(n+1) + (-1)^{n+1}}{n+1}} + \dfrac{(-1)^{n+2}(2n+3)}{(n+1)(n+2)}$

$\dfrac{(n+1)(n+2) + (-1)^{n+1}(n+2)}{(n+1)(n+2)} + \dfrac{(-1)^{n+2}(2n+3)}{(n+1)(n+2)}$

$\dfrac{(n+1)(n+2) + (-1)^{n+1}(n+2) - (-1)^{n+1}(2n+3) }{(n+1)(n+2)}$

$\dfrac{(n+1)(n+2) + (-1)^{n+1}[(n+2) - (2n+3)] }{(n+1)(n+2)}$

$\dfrac{(n+1)(n+2) + (-1)^{n+2}(n+1) }{(n+1)(n+2)}$

$\dfrac{(n+1)(n+2)}{(n+1)(n+2)}+ \dfrac{(-1)^{n+2}(n+1)}{(n+1)(n+2)}$

$ 1 + \dfrac{(-1)^{(n+1)+1}}{(n+1)+1}$
 

Related to Proof with Induction 3/2-5/6+7/12-9/20+11/30-....

1. What is proof by induction?

Proof by induction is a mathematical method used to prove that a statement or proposition is true for all natural numbers. It involves using a base case and an inductive step to show that the statement holds for all subsequent numbers.

2. How does proof by induction work?

Proof by induction works by first proving that a statement is true for a specific base case, typically the smallest natural number. Then, the inductive step is used to show that if the statement holds for one number, it also holds for the next number. This process is repeated until it can be shown that the statement is true for all natural numbers.

3. What is the purpose of using proof by induction?

The purpose of using proof by induction is to prove that a statement or proposition is true for all natural numbers. This can be useful in mathematics and other fields to prove the validity of a theory or formula.

4. How is proof by induction different from other proof methods?

Proof by induction differs from other proof methods in that it specifically focuses on proving statements for all natural numbers. Other proof methods may focus on specific cases or use different logical reasoning to prove a statement.

5. How is proof by induction applied to the series 3/2-5/6+7/12-9/20+11/30-....?

In this series, proof by induction would be used to show that the statement "the sum of the first n terms of the series is equal to (2n-1)/2n" is true for all natural numbers n. The base case would be n=1, where the sum is equal to 3/2. Then, the inductive step would be used to show that if the statement holds for n, it also holds for n+1. This would be repeated until it can be shown that the statement is true for all natural numbers.

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