[Editor: Find a closed formula for?]
$$1 + 2^4x + 3^4x^2 + 4^4x^3 + \dots+n^4{x^{n-1}}+\dots$$
I think Vn method should be applied but can't seem to figure it out.
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1The image is unnecessary. Even unformatted text is better than an image link, if the meaning is clear. Links can go away, images can't be read by screen readers, and images won't show up in searches for a formula. – aschepler Oct 03 '23 at 14:07
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The original image is of an infinite series, however the body is about a finite sum. Can you clarify which is intended? – B. Mehta Oct 03 '23 at 14:18
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@B.Mehta my bad, i fixed it. – shivansh tiwari Oct 03 '23 at 14:23
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Is the last term supposed to be $5^5$? – jjagmath Oct 03 '23 at 14:33
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Sart from $\sum_{n=0}^\infty x^n=\frac1{1-x}$ (for $|x|<1$). Differentiate and multiply by $x$, 4 times. This must be a duplicate but I haven't been able to find it. – Anne Bauval Oct 03 '23 at 14:47
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@jjagmath last term? its an infinite series... if I misunderstood please correct me – shivansh tiwari Oct 03 '23 at 14:49
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@shivanshtiwari You did a typo in your last edit (you wrote $1 + 2^4x + 3^4x^2 + 4^4x^3 + 5^{\color{red}5}x^4 ...$), I fixed it. – Anne Bauval Oct 03 '23 at 14:50
1 Answers
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just follow Anne Bauval's comment:
$$\sum_{n=0}^{\infty} x^n = \frac1{1-x} \implies \sum_{n=0}^{\infty} nx^{n-1} = \frac1{(1-x)^2} \implies \sum_{n=0}^{\infty} nx^{n} = \frac x{(1-x)^2} \implies\\ \implies \sum_{n=0}^{\infty} n^2x^{n-1} = \frac {x+1}{(1-x)^3} \implies \sum_{n=0}^{\infty} n^2x^{n} = x \cdot\frac {x+1}{(1-x)^3} \implies \\ \implies \sum_{n=0}^{\infty} n^3x^{n-1} = \frac {x^2+4x+1}{(1-x)^3} \implies \sum_{n=0}^{\infty} n^3x^{n} = x \cdot \frac {x^2+4x+1}{(1-x)^3} \implies \\ \implies \sum_{n=0}^{\infty} n^4x^{n-1} = \boxed{\frac {x^3+11x^2+11x+1}{(1-x)^4} } = \frac {(x+1)(x^2+10x+1)}{(1-x)^4} $$
hellofriends
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