Given 1⋅21+2⋅22+3⋅23+4⋅24+…+d⋅2d=∑r=1dr⋅2r how can we infer to the following solution? 2(d−1)⋅2d+2
how can we infer to the following solution?
Answer & Explanation
Beginner2022-01-22Added 16 answers
As noted above, observe that
Differentiating both sides and multiplying by x, we find
Beginner2022-01-23Added 12 answers
You can use iterative integration of finite differences. The method works as follows. Assume your problem statement is: Assume there is a function such that: By finite difference we have: Now assume we have a guess for such that some error remains: The error itself can be expressed as a new integral since finite difference distributes over sums. We have: And hence we have a new sub problem: Lets apply the method to the problem at hand, we have: We can guess: We arrive at: We can guess again: We arrive at: So the integration terminated, and the closed form integral is: Using this for the sum we get:
Skilled2022-01-27Added 437 answers
Summation by parts gives that with the choice we have
and the RHS simplifies to as wanted.
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