[Largedisplaystyle sum_{n=1}^{infty}frac{H_{2n}}{n(6n+1)}
]
(Largemathbf{Solution:})
Let (S) denote the sum. Then
[egin{align*}
S=sum_{n=1}^infty frac{H_{2n}}{n(6n+1)} &= sum_{n=1}^inftyfrac{H_{2n}}{n}int_0^1 x^{6n}mathrm dx \
&= int_0^1left( sum_{n=1}^inftyfrac{H_{2n}}{n}x^{6n}
ight)mathrm dx ag{1}end{align*}]
Let (displaystyle f(x)=sum_{n=1}^infty frac{H_n}{n}x^n) where (|x|<1). It can be shown that
[egin{align*}
f(x)= ext{Li}_2(x)+frac{1}{2}ln^2(1-x) ag{2}
end{align*}]
Then, we can write
[egin{align*}
sum_{n=1}^inftyfrac{H_{2n}}{n}x^{6n} &= fleft(x^3
ight)+fleft(-x^3
ight) \
&= ext{Li}_2left(x^3
ight)+ ext{Li}_2left(-x^3
ight)+frac{ln^2left(1-x^3
ight)+ln^2left(1+x^3
ight)}{2} ag{3}
end{align*}]
Substitute (3) into (1) to get
[egin{align*}
S=int_0^1 left( ext{Li}_2left(x^3
ight)+ ext{Li}_2left(-x^3
ight)+frac{ln^2left(1-x^3
ight)+ln^2left(1+x^3
ight)}{2}
ight)mathrm dx ag{4}
end{align*}]
Note that
[egin{align*}
int_0^1left( ext{Li}_2left(x^3
ight)+ ext{Li}_2left(-x^3
ight)
ight)mathrm dx &= frac{1}{2}int_0^1sum_{n=1}^inftyfrac{x^{6n}}{n^2} mathrm dx \
&=frac{1}{2}sum_{n=1}^inftyfrac{1}{n^2(6n+1)} \
&= frac{1}{2}sum_{n=1}^infty left(frac{1}{n^2}-frac{6}{n}+frac{36}{1+6n}
ight) \
&= frac{1}{2}left(frac{pi^2}{6} -6psi_0left(frac{1}{6}
ight)-6gamma_0-36
ight) \
&= frac{pi^2}{12}+frac{3pisqrt{3}}{2}+6ln 2+frac{9}{2}ln 3-18 ag{5}
end{align*}]
[egin{align*}
frac{1}{2}int_0^1ln^3left(1-x^3
ight)mathrm dx &= frac{1}{6}int_0^1t^{-2/3}ln^2(1-t)mathrm dt quad (t=x^3)\
&= frac{1}{6}left[frac{partial^2}{partial y^2} mathrm{B}(x,y)
ight]_{x=1/3,y=1} \
&= frac{pi ^2}{8}-frac{sqrt{3} pi }{2}+frac{9}{2}+frac{9}{8} ln^2 3-frac{9 ln 3}{2}+frac{1}{4} sqrt{3} pi ln 3-frac{psi_1left(dfrac{4}{3}
ight)}{2} ag{6}
end{align*}]
Substitute (5) and (6) into equation (4) to get
[S=-frac{27}{2}+frac{5pi^2}{24}+frac{9}{8}ln^2 3+frac{pisqrt{3}}{4}(4+ln 3)+6ln 2-frac{1}{2}psi_1left(frac{4}{3}
ight)+frac{1}{2}int_0^1 ln^2(1+x^3)mathrm{d}x
]
Now, it remains to calculate (displaystyle int_0^1 ln^2(1+x^3)mathrm{d}x).
According to Mathematica, it equals
[egin{align*}
int_0^1 ln^2(1+x^3)mathrm{d}x&=18-frac{5}{36}pi ^{2}+frac{ln^{2}3}{4}+3ln^{2}2-12ln 2+frac{lndfrac{2187}{16}-12}{2sqrt{3}}pi +mathrm{Li}_{2}left ( -frac{1}{3}
ight )\
&~~~-left ( 1+isqrt{3}
ight )mathrm{Li}_{2}left ( frac{3-isqrt{3}}{6}
ight )+left ( 1-isqrt{3}
ight )mathrm{Li}_{2}left ( frac{3-isqrt{3}}{4}
ight )\
&~~~-left ( 1-isqrt{3}
ight )mathrm{Li}_{2}left ( frac{3+isqrt{3}}{6}
ight )+left ( 1+isqrt{3}
ight )mathrm{Li}_{2}left ( frac{3+isqrt{3}}{4}
ight )
end{align*}]
Hence, the final result is
[oxed{displaystyle egin{align*}
sum_{n=1}^{infty}frac{H_{2n}}{n(6n+1)}&=color{blue}{-frac{9}{2}+frac{5}{36}pi ^{2} +frac{5}{4}ln^23+frac{3}{2}ln^22-frac{ln 2}{sqrt{3}}+left ( frac{7}{4sqrt{3}}+frac{sqrt{3}}{4}
ight )ln 3}\
&~~~color{blue}{+frac{1}{2}Bigg{mathrm{Li}_{2}left ( -frac{1}{3}
ight )-psi _{1}left ( frac{4}{3}
ight )-left ( 1+isqrt{3}
ight )mathrm{Li}_{2}left ( frac{3-isqrt{3}}{6}
ight )}\
&~~~color{blue}{+left ( 1-isqrt{3}
ight )mathrm{Li}_{2}left ( frac{3-isqrt{3}}{4}
ight )-left ( 1-isqrt{3}
ight )mathrm{Li}_{2}left ( frac{3+isqrt{3}}{6}
ight )}\
&~~~color{blue}{+left ( 1+isqrt{3}
ight )mathrm{Li}_{2}left ( frac{3+isqrt{3}}{4}
ight )Bigg}}
end{align*}}]