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print "hello world"
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Math
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2 \times 3
2×3
2^{34}
234
a_{i-1}
ai−1
\frac{2}{3}
32
\sqrt{2}
2
\sum_{i=1}^3
∑i=13
\sin \theta
sinθ
\boxed{123}
123
Comments
n=1∑∞n!n2======n=1∑∞n!n2−n+nn=1∑∞n!n(n−1)+n=1∑∞n!nn=2∑∞(n−1)!n−1+n=1∑∞(n−1)!1n=2∑∞(n−2)!1+n=1∑∞(n−1)!1, let p=n−2, let q=n−1p=0∑∞p!1+q=0∑∞q!1e+e=2e
The trick to evaluate n=1∑∞n!nk for a positive integer k is to determine a linear combination of nk in terms of n,n(n−1),n(n−1)(n−2),…,n(n−1)(n−2)⋅⋅⋅(n−k+1)
For example n4=n(n−1)(n−2)(n−3)+6n(n−1)(n−2)+7n(n−1)+n
Easy Math Editor
This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
When posting on Brilliant:
*italics*or_italics_**bold**or__bold__paragraph 1
paragraph 2
[example link](https://brilliant.org)> This is a quote# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"\(...\)or\[...\]to ensure proper formatting.2 \times 32^{34}a_{i-1}\frac{2}{3}\sqrt{2}\sum_{i=1}^3\sin \theta\boxed{123}Comments
n=1∑∞n!n2======n=1∑∞n!n2−n+nn=1∑∞n!n(n−1)+n=1∑∞n!nn=2∑∞(n−1)!n−1+n=1∑∞(n−1)!1n=2∑∞(n−2)!1+n=1∑∞(n−1)!1, let p=n−2, let q=n−1p=0∑∞p!1+q=0∑∞q!1e+e=2e
The trick to evaluate n=1∑∞n!nk for a positive integer k is to determine a linear combination of nk in terms of n, n(n−1), n(n−1)(n−2), … , n(n−1)(n−2)⋅⋅⋅(n−k+1)
For example n4=n(n−1)(n−2)(n−3)+6n(n−1)(n−2)+7n(n−1)+n
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Why did you change the limits? It was from n=1..you made it n=2..
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Because (n−1)!n−1=0 when n=1
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Oh, right..sorry.
Great! Thanks.
e^x =1+ x + x^2/2.1 + x^3/3.2.1 + ..... up to infinity =1 + 2x/2.1+ 3x^2/3.2.1 + 4x^3/4.3.2.1 +.....
x.e^x = x + 2x^2/2.1 + 3x^3/3.2.1 + 4x^4/4.3.2.1 + ....... ------- (1)
differentiating eq. (1) w.r.t. x we get the above series. Then substituting x=1 we get the answer as 2e .
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Nicely done!