I want to calculate this sum. How can I do it ? $$ \sum_{i=1}^n i \cdot \left(\frac{9}{10}\right)^i$$ I know how to calc geometric sum, but how to calc this?
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See this. – Git Gud May 20 '14 at 10:35
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See a related one: https://math.stackexchange.com/questions/250746/calculate-the-sum-of-series-sum-limits-i-0n-1-i2i – gar May 20 '14 at 10:35
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If you take into account the other comments, it is the derivative of a geometric sum. Then ... – Claude Leibovici May 20 '14 at 10:42
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Let x = 0.9
The sum is $\displaystyle x + 2x^2 + 3x^3 + ... + nx^n \\ = x(1 + 2x + 3x^2 + ... nx^{n-1}) \\ = x.\frac{d}{dx}(x + x^2 + x^3 + ... + x^n) \\ = x.\frac{d}{dx}(\frac{x(x^n-1)}{x-1})$
Evaluate the derivative by quotient rule and simplify the expression to give a general sum. Now you can put 0.9 back into it. ;)
Another way: call $\displaystyle x + 2x^2 + 3x^3 + ... + nx^n = S_n(x)\\ xS_n(x) = x^2 + 2x^3 + ... + (n-1)x^n + nx^{n+1}$
Subtracting, $\displaystyle (1-x)S_n(x)= x + x^2 + x^3 + ... + x^n - nx^{n+1} = \frac{x(x^n-1)}{x-1} - nx^{n+1}$
The rest is just algebra. This way you avoid calculus.
Deepak
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