Mathematica Stack Exchange is a question and answer site for users of Mathematica. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I want to solve the equation

$$ 2n\cdot\binom{n}{0} + 5(n-1)\binom{n}{1}+13(n-2)\binom{n}{2}+\cdots + \left(2^{n-1} + 3^{n-1}\right)\binom{n}{n-1} = 1685. $$

or using alternative notion for binomial coefficients:

$$ 2nC_{n}^0 + 5(n-1)C_n^1 + 13(n-2)C_n^2 +\cdots + \left(2^{n-1} + 3^{n-1}\right) C_n^{n-1}=1685. $$

where, $n$ is a natural number. I do not know how to solve this. How do I tell Mathematica to do that?

share|improve this question
up vote 13 down vote accepted

For the summation, Mathematica can obtain a closed form, then we can use Reduce or FindInstance to get the answer:

Sum[(2^i + 3^i)*(n - i)*Binomial[n, i], {i, 0, n - 1}]

(1/12)*(4*3^n + 3*4^n)*n

Reduce[% == 1685, n, Integers]

n == 5

share|improve this answer

You can try this

    Sum[(2^i + 3^i) (n - i) Binomial[n, i], {i, 0, k}] == 1685, n, 
  Integers], {k, 1, 50, 1}]
share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.