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Finger [1]
2 years ago
10

1+4=5, 2+5=12, 3+6=21, 8+11?

Mathematics
1 answer:
Mrrafil [7]2 years ago
7 0
The answer is 96.
Solution:
1+4(5) = 5
2+5(7) = 12(5+7)
3+6(9) = 21(12+9)
4+7(11) = 32(21+11)
5+8(13) = 45(32+13)
6+9(15) = 60(45+15)
7+10(17) = 77(60+17)
8+11(19) = 96(77+19).
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Find a compact form for generating functions of the sequence 1, 8,27,... , k^3
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This sequence has generating function

F(x)=\displaystyle\sum_{k\ge0}k^3x^k

(if we include k=0 for a moment)

Recall that for |x|, we have

\displaystyle\frac1{1-x}=\sum_{k\ge0}x^k

Take the derivative to get

\displaystyle\frac1{(1-x)^2}=\sum_{k\ge0}kx^{k-1}=\frac1x\sum_{k\ge0}kx^k

\implies\dfrac x{(1-x)^2}=\displaystyle\sum_{k\ge0}kx^k

Take the derivative again:

\displaystyle\frac{(1-x)^2+2x(1-x)}{(1-x)^4}=\sum_{k\ge0}k^2x^{k-1}=\frac1x\sum_{k\ge0}k^2x^k

\implies\displaystyle\frac{x+x^2}{(1-x)^3}=\sum_{k\ge0}k^2x^k

Take the derivative one more time:

\displaystyle\frac{(1+2x)(1-x)^3+3(x+x^2)(1-x)^2}{(1-x)^6}=\sum_{k\ge0}k^3x^{k-1}=\frac1x\sum_{k\ge0}k^3x^k

\implies\displaystyle\frac{x+4x^3+x^3}{(1-x)^4}=\sum_{k\ge0}k^3x^k

so we have

\boxed{F(x)=\dfrac{x+4x^3+x^3}{(1-x)^4}}

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