Figurate number representing a decagon
Inmathematics , adecagonal number is afigurate number that extends the concept oftriangular andsquare numbers to thedecagon (a ten-sided polygon).[ 1] n -th decagonal numbers counts the dots in a pattern ofn nested decagons, all sharing a common corner, where thei th decagon in the pattern has sides made ofi dots spaced one unit apart from each other. Then -th decagonal number is given by the following formula
d n = 4 n 2 − 3 n {\displaystyle d_{n}=4n^{2}-3n} [ 2] The first few decagonal numbers are:
0 ,1 ,10 ,27 ,52 ,85 ,126 ,175 ,232 ,297 , 370, 451, 540, 637, 742, 855, 976,1105 , 1242, 1387, 1540, 1701, 1870, 2047, 2232, 2425, 2626, 2835, 3052, 3277, 3510, 3751,4000 , 4257, 4522, 4795, 5076, 5365, 5662, 5967, 6280, 6601, 6930, 7267, 7612, 7965, 8326 (sequenceA001107 OEIS ).Then th decagonal number can also be calculated by adding the square ofn to thrice the (n −1)thpronic number or, to put it algebraically, as
D n = n 2 + 3 ( n 2 − n ) {\displaystyle D_{n}=n^{2}+3\left(n^{2}-n\right)} D n = D n − 1 + 8 n − 7 , D 0 = 0 {\displaystyle D_{n}=D_{n-1}+8n-7,D_{0}=0} D n = 2 D n − 1 − D n − 2 + 8 , D 0 = 0 , D 1 = 1 {\displaystyle D_{n}=2D_{n-1}-D_{n-2}+8,D_{0}=0,D_{1}=1} D n = 3 D n − 1 − 3 D n − 2 + D n − 3 , D 0 = 0 , D 1 = 1 , D 2 = 10 {\displaystyle D_{n}=3D_{n-1}-3D_{n-2}+D_{n-3},D_{0}=0,D_{1}=1,D_{2}=10} Thesum of the reciprocals of the decagonal numbers admits a simple closed form:∑ n = 1 ∞ 1 4 n 2 − 3 n + ∑ n = 1 ∞ 1 n ( 4 n − 3 ) = ln ( 2 ) + π 6 . {\displaystyle \sum _{n=1}^{\infty }{\frac {1}{4n^{2}-3n}}+\sum _{n=1}^{\infty }{\frac {1}{n\left(4n-3\right)}}=\ln \left(2\right)+{\frac {\pi }{6}}.}
This derivation rests upon the method of adding a "constructive zero":∑ n = 1 ∞ 1 n ( 4 n − 3 ) = 4 3 ∑ n = 1 ∞ ( 1 4 n − 3 − 1 4 n ) = 2 3 ∑ n = 1 ∞ ( 2 4 n − 3 − 2 4 n + ( 1 4 n − 1 − 1 4 n − 2 ) − ( 1 4 n − 1 − 1 4 n − 2 ) ) {\displaystyle {\begin{aligned}\sum _{n=1}^{\infty }{\frac {1}{n\left(4n-3\right)}}&{}={\frac {4}{3}}\sum _{n=1}^{\infty }\left({\frac {1}{4n-3}}-{\frac {1}{4n}}\right)\\&={\frac {2}{3}}\sum _{n=1}^{\infty }\left({\frac {2}{4n-3}}-{\frac {2}{4n}}+\left({\frac {1}{4n-1}}-{\frac {1}{4n-2}}\right)-\left({\frac {1}{4n-1}}-{\frac {1}{4n-2}}\right)\right)\end{aligned}}} = 2 3 ∑ n = 1 ∞ [ ( 1 4 n − 3 − 1 4 n − 2 + 1 4 n − 1 − 1 4 n ) + ( 1 4 n − 2 − 1 4 n ) + ( 1 4 n − 3 − 1 4 n − 1 ) ] = 2 3 ∑ n = 1 ∞ ( 1 4 n − 3 − 1 4 n − 2 + 1 4 n − 1 − 1 4 n ) + 1 3 ∑ n = 1 ∞ ( 1 2 n − 1 − 1 2 n ) + 2 3 ∑ n = 1 ∞ ( 1 2 ( 2 n − 1 ) − 1 − 1 2 ( 2 n ) − 1 ) = 2 3 ∑ n = 1 ∞ ( − 1 ) n + 1 n + 1 3 ∑ n = 1 ∞ ( − 1 ) n + 1 n + 2 3 ∑ n = 1 ∞ ( − 1 ) n + 1 2 n − 1 = ln ( 2 ) + π 6 . {\displaystyle {\begin{aligned}&={\frac {2}{3}}\sum _{n=1}^{\infty }\left[\left({\frac {1}{4n-3}}-{\frac {1}{4n-2}}+{\frac {1}{4n-1}}-{\frac {1}{4n}}\right)+\left({\frac {1}{4n-2}}-{\frac {1}{4n}}\right)+\left({\frac {1}{4n-3}}-{\frac {1}{4n-1}}\right)\right]\\&={\frac {2}{3}}\sum _{n=1}^{\infty }\left({\frac {1}{4n-3}}-{\frac {1}{4n-2}}+{\frac {1}{4n-1}}-{\frac {1}{4n}}\right)+{\frac {1}{3}}\sum _{n=1}^{\infty }\left({\frac {1}{2n-1}}-{\frac {1}{2n}}\right)+{\frac {2}{3}}\sum _{n=1}^{\infty }\left({\frac {1}{2(2n-1)-1}}-{\frac {1}{2(2n)-1}}\right)\\&={\frac {2}{3}}\sum _{n=1}^{\infty }{\frac {(-1)^{n+1}}{n}}+{\frac {1}{3}}\sum _{n=1}^{\infty }{\frac {(-1)^{n+1}}{n}}+{\frac {2}{3}}\sum _{n=1}^{\infty }{\frac {(-1)^{n+1}}{2n-1}}\\&=\ln \left(2\right)+{\frac {\pi }{6}}.\end{aligned}}}
Possessing a specific set of other numbers
Expressible via specific sums
