Proving 𝛑² = n(n+1) 2ⁿ⁻² w/ Clever Trick

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In summary, the conversation discusses a clever way to prove the identity \sum_{i=1}^{n}i^2 {n \choose i} = n(n+1) 2^{n-2} using the second derivative of (1+x)^n and its binomial expansion. The speaker initially mistakenly replaces (n+1) with (n-1) but then corrects themselves with the identity \sum_{i=0}^n i(i-1) {n \choose i} = n(n-1)2^{n-2}.
  • #1
ehrenfest
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Homework Statement


Does anyone know a clever way to prove that

[tex] \sum_{i=1}^{n}i^2 {n \choose i} = n(n+1) 2^{n-2} [/tex]

where B(n,i) is n take i?

I can do it, but I had to divide into the cases of n = odd and n = even and it took about 1 page front and back. I'm sure there is a trick.

Homework Equations


The Attempt at a Solution

 
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  • #2
Take the second derivative of (1+x)^n and its binomial expansion, then mess with the index of the summation and set x=1. (I'm assuming your n+1 is actually n-1.)
 
  • #3
morphism said:
Take the second derivative of (1+x)^n and its binomial expansion, then mess with the index of the summation and set x=1. (I'm assuming your n+1 is actually n-1.)
? It's obviously not true if you replace (n+1) by (n-1). Take n= 1. Then the lefthand side is 1. [iotex]n(n+1)2^{n-2}[/itex] becomes, for n= 1, [itex]1(2)2^{-1}= 1[/itex]. If you replace (n+1) by (n-1), it becomes [tex]2(0)2^{-1}= 0[/itex].
 
  • #4
You're right of course. I was a bit careless with my algebra. The identity I had in mind was:
[tex]\sum_{i=0}^n i(i-1) {n \choose i} = n(n-1)2^{n-2}[/tex]

But no worries - a similar trick can still be applied: Differentiate, multiply by x, and differentiate again!
 

FAQ: Proving 𝛑² = n(n+1) 2ⁿ⁻² w/ Clever Trick

How can you prove the equation 𝛑² = n(n+1) 2ⁿ⁻² with a clever trick?

The proof for this equation involves using a clever method known as mathematical induction. This method involves proving a statement for the first value of n, and then showing that if it is true for any arbitrary value, it must also be true for the next value of n. By repeating this process, the statement can be proven for all values of n, including infinity.

What is the significance of proving 𝛑² = n(n+1) 2ⁿ⁻²?

This equation is significant as it is a fundamental result in the field of mathematics. It shows the relationship between the area of a circle and its radius, and is used in many mathematical and scientific calculations involving circles.

Can you explain the steps involved in proving 𝛑² = n(n+1) 2ⁿ⁻² with mathematical induction?

The steps involved in this proof include proving the statement for the base case (n=1), assuming it is true for an arbitrary value of n (known as the inductive hypothesis), and then using this assumption to prove the statement for the next value of n. This process is repeated until the statement is proven for all values of n.

Are there any alternative methods for proving 𝛑² = n(n+1) 2ⁿ⁻²?

Yes, there are alternative methods for proving this equation, such as using geometric or algebraic proofs. However, using mathematical induction is often the most straightforward and efficient method.

What are some real-world applications of the equation 𝛑² = n(n+1) 2ⁿ⁻²?

This equation has many practical applications, including in geometry, physics, and engineering. For example, it is used to calculate the area of a circle, the volume of a sphere, and the circumference of a circle. It also has applications in calculating the force of gravity between two objects and in determining the amount of space needed for computer memory storage.

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