5
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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?

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13
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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

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3
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You can try this

Manipulate[
 Solve[FunctionExpand@
    Sum[(2^i + 3^i) (n - i) Binomial[n, i], {i, 0, k}] == 1685, n, 
  Integers], {k, 1, 50, 1}]
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