Closed Form of an Infinite Series

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SUMMARY

The closed form for the infinite series 1*C(n,1) + 2*C(n,2) + 3*C(n,3) + ... + n*C(n,n) can be derived using the binomial theorem and differentiation. Specifically, differentiating the binomial expansion (1+x)^n and evaluating at x=1 leads to the result n*2^(n-1). The series is not infinite in the traditional sense, as it sums only up to n terms, where C(n,k) represents the binomial coefficient.

PREREQUISITES
  • Understanding of binomial coefficients, specifically C(n,k) = n!/(k!(n-k)!)
  • Familiarity with the binomial theorem and its applications
  • Basic calculus concepts, particularly differentiation
  • Knowledge of series and summation notation
NEXT STEPS
  • Study the binomial theorem and its implications in combinatorics
  • Learn about differentiation techniques and their applications in series
  • Explore properties of binomial coefficients and their identities
  • Investigate other series summation techniques and closed forms
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Mathematics students, educators, and anyone interested in combinatorial identities and series summation techniques.

Frillth
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Homework Statement



I'm looking to find a closed form for the infinite series:
1*C(n,1) + 2*C(n,2) + 3*C(n,3) + ... + n*C(n,n)

Homework Equations



C(n,k) = n!/(k!*(n-k)!)
C(n,1) + C(n,2) + C(n,3) + ... + C(n,n) = 2^n - 1

The Attempt at a Solution



I'm not quite sure where to start this problem. Any tips?
 
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(1+x)^n=C(0,n)+C(1,n)*x+C(2,n)*x^2+...+C(n,n)*x^n. This is why it's clear that your second identity is pretty obvious (put x=1). Think what happens when you differentiate that wrt x and then put x=1. Oh, and your series really isn't that infinite, is it?
 
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