Return to Answer

Not an answer, but a reduction to a definite integral.

First, Lagrange inversion implies that the generating function for $T_k$ is $${\cal T}(z):=(1-2z-3z^2)^{-\frac12}=((1+z)(1-3z))^{-\frac12},$$ and thus $T_k = [z^k]\ {\cal T}(z)$. Notice that $${\cal T}'(z) = \frac{1+3z}{(1+z)(1-3z)}{\cal T}(z).$$

Second, by the property of the beta function, we have $$\frac{1}{k\binom{2k}k} = B(k+1,k) = \int_0^1 x^k(1-x)^{k-1}\,{\rm d}x = \frac12 \int_0^1 x^{k-1}(1-x)^{k-1}\,{\rm d}x.$$

It follows that $$I_1:=\sum_{k\geq 1} \frac{T_{k-1}}{k^2\binom{2k}k 3^{k-1}} = \frac14 \int_0^1\int_0^1 {\cal T}\big( \frac{x(1-x)y(1-y)}{3} \big)\,{\rm d}x{\rm d}y$$ and $$I_2 := \sum_{k\geq 1} \frac{(k-1) T_{k-1}}{k^2\binom{2k}k 3^{k-1}} = \frac14 \int_0^1\int_0^1\, \frac{x(1-x)y(1-y)}3\, {\cal T}'\big( \frac{x(1-x)y(1-y)}{3} \big)\,{\rm d}x{\rm d}y.$$

Since $105k-44 = 105(k-1)+61$, the first sum in question equals $105 I_2 + 61 I_1$.

The second sum in question can be reduced to a definite integral similarly.

Not an answer, but a reduction to a definite integral.

First, Lagrange inversion implies that the generating function for $T_k$ is $${\cal T}(z):=(1-2z-3z^2)^{-\frac12}=((1+z)(1-3z))^{-\frac12},$$ and thus $T_k = [z^k]\ {\cal T}(z)$. Notice that $${\cal T}'(z) = \frac{1+3z}{(1+z)(1-3z)}{\cal T}(z).$$

Second, by the property of the beta function, we have $$\frac{1}{k\binom{2k}k} = B(k+1,k) = \int_0^1 x^k(1-x)^{k-1}\,{\rm d}x = \frac12 \int_0^1 x^{k-1}(1-x)^{k-1}\,{\rm d}x.$$

It follows that $$I_1:=\sum_{k\geq 1} \frac{T_{k-1}}{k^2\binom{2k}k^2 3^{k-1}} = \frac14 \int_0^1\int_0^1 {\cal T}\big( \frac{x(1-x)y(1-y)}{3} \big)\,{\rm d}x{\rm d}y$$ and $$I_2 := \sum_{k\geq 1} \frac{(k-1) T_{k-1}}{k^2\binom{2k}k^2 3^{k-1}} = \frac14 \int_0^1\int_0^1\, \frac{x(1-x)y(1-y)}3\, {\cal T}'\big( \frac{x(1-x)y(1-y)}{3} \big)\,{\rm d}x{\rm d}y.$$

Since $105k-44 = 105(k-1)+61$, the first sum in question equals $105 I_2 + 61 I_1$.

The second sum in question can be reduced to a definite integral similarly.

Not an answer, but a reduction to a definite integral.

First, Lagrange inversion implies that the generating function for $T_k$ is $${\cal T}(z):=(1-2z-3z^2)^{-\frac12}=((1+z)(1-3z))^{-\frac12},$$ and thus $T_k = [z^k]\ {\cal T}(z)$. Notice that $${\cal T}'(z) = \frac{1+3z}{(1+z)(1-3z)}{\cal T}(z).$$

Second, by the property of the beta function, we have $$\frac{1}{k\binom{2k}k} = B(k+1,k) = \int_0^1 x^k(1-x)^{k-1}\,{\rm d}x.$$

It follows that $$I_1:=\sum_{k\geq 1} \frac{T_{k-1}}{k^2\binom{2k}k 3^{k-1}} = \int_0^1\int_0^1 xy\, {\cal T}\big( \frac{x(1-x)y(1-y)}{3} \big)\,{\rm d}x{\rm d}y$$ and $$I_2 := \sum_{k\geq 1} \frac{(k-1) T_{k-1}}{k^2\binom{2k}k 3^{k-1}} = \int_0^1\int_0^1\, \frac{x^2y^2(1-x)(1-y)}3\, {\cal T}'\big( \frac{x(1-x)y(1-y)}{3} \big)\,{\rm d}x{\rm d}y.$$

Since $105k-44 = 105(k-1)+61$, the first sum in question equals $105 I_2 + 61 I_1$.

The second sum in question can be reduced to a definite integral similarly.

P.S. By symmetry between $x$ and $1-x$, and between $y$ and $1-y$, we also have $$I_1 = \frac14 \int_0^1\int_0^1 {\cal T}\big( \frac{x(1-x)y(1-y)}{3} \big)\,{\rm d}x{\rm d}y$$ and $$I_2 = \frac14 \int_0^1\int_0^1\, \frac{x(1-x)y(1-y)}3\, {\cal T}'\big( \frac{x(1-x)y(1-y)}{3} \big)\,{\rm d}x{\rm d}y.$$

Not an answer, but a reduction to a definite integral.

First, Lagrange inversion implies that the generating function for $T_k$ is $${\cal T}(z):=(1-2z-3z^2)^{-\frac12}=((1+z)(1-3z))^{-\frac12},$$ and thus $T_k = [z^k]\ {\cal T}(z)$. Notice that $${\cal T}'(z) = \frac{1+3z}{(1+z)(1-3z)}{\cal T}(z).$$

Second, by the property of the beta function, we have $$\frac{1}{k\binom{2k}k} = B(k+1,k) = \int_0^1 x^k(1-x)^{k-1}\,{\rm d}x = \frac12 \int_0^1 x^{k-1}(1-x)^{k-1}\,{\rm d}x.$$

It follows that $$I_1:=\sum_{k\geq 1} \frac{T_{k-1}}{k^2\binom{2k}k 3^{k-1}} = \frac14 \int_0^1\int_0^1 {\cal T}\big( \frac{x(1-x)y(1-y)}{3} \big)\,{\rm d}x{\rm d}y$$ and $$I_2 := \sum_{k\geq 1} \frac{(k-1) T_{k-1}}{k^2\binom{2k}k 3^{k-1}} = \frac14 \int_0^1\int_0^1\, \frac{x(1-x)y(1-y)}3\, {\cal T}'\big( \frac{x(1-x)y(1-y)}{3} \big)\,{\rm d}x{\rm d}y.$$

Since $105k-44 = 105(k-1)+61$, the first sum in question equals $105 I_2 + 61 I_1$.

The second sum in question can be reduced to a definite integral similarly.

Not an answer, but a reduction to a definite integral.

First, Lagrange inversion implies that the generating function for $T_k$ is $${\cal T}(z):=(1-2z-3z^2)^{-\frac12}=((1+z)(1-3z))^{-\frac12},$$ and thus $T_k = [z^k]\ {\cal T}(z)$. Notice that $${\cal T}'(z) = \frac{1+3z}{(1+z)(1-3z)}{\cal T}(z).$$

Second, we have $$\frac{1}{k\binom{2k}k} = \int_0^1 x^k(1-x)^{k-1}\,{\rm d}x.$$

It follows that $$I_1:=\sum_{k\geq 1} \frac{T_{k-1}}{k^2\binom{2k}k 3^{k-1}} = \int_0^1\int_0^1 xy\, {\cal T}\big( \frac{x(1-x)y(1-y)}{3} \big)\,{\rm d}x{\rm d}y$$ and $$I_2 := \sum_{k\geq 1} \frac{(k-1) T_{k-1}}{k^2\binom{2k}k 3^{k-1}} = \int_0^1\int_0^1\, \frac{x^2y^2(1-x)(1-y)}3\, {\cal T}'\big( \frac{x(1-x)y(1-y)}{3} \big)\,{\rm d}x{\rm d}y.$$

Since $105k-44 = 105(k-1)+61$, the first sum in question equals $105 I_2 + 61 I_1$.

The second sum in question can be reduced to a definite integral similarly.

Not an answer, but a reduction to a definite integral.

First, Lagrange inversion implies that the generating function for $T_k$ is $${\cal T}(z):=(1-2z-3z^2)^{-\frac12}=((1+z)(1-3z))^{-\frac12},$$ and thus $T_k = [z^k]\ {\cal T}(z)$. Notice that $${\cal T}'(z) = \frac{1+3z}{(1+z)(1-3z)}{\cal T}(z).$$

Second, by the property of the beta function, we have $$\frac{1}{k\binom{2k}k} = B(k+1,k) = \int_0^1 x^k(1-x)^{k-1}\,{\rm d}x.$$

It follows that $$I_1:=\sum_{k\geq 1} \frac{T_{k-1}}{k^2\binom{2k}k 3^{k-1}} = \int_0^1\int_0^1 xy\, {\cal T}\big( \frac{x(1-x)y(1-y)}{3} \big)\,{\rm d}x{\rm d}y$$ and $$I_2 := \sum_{k\geq 1} \frac{(k-1) T_{k-1}}{k^2\binom{2k}k 3^{k-1}} = \int_0^1\int_0^1\, \frac{x^2y^2(1-x)(1-y)}3\, {\cal T}'\big( \frac{x(1-x)y(1-y)}{3} \big)\,{\rm d}x{\rm d}y.$$

Since $105k-44 = 105(k-1)+61$, the first sum in question equals $105 I_2 + 61 I_1$.

The second sum in question can be reduced to a definite integral similarly.

P.S. By symmetry between $x$ and $1-x$, and between $y$ and $1-y$, we also have $$I_1 = \frac14 \int_0^1\int_0^1 {\cal T}\big( \frac{x(1-x)y(1-y)}{3} \big)\,{\rm d}x{\rm d}y$$ and $$I_2 = \frac14 \int_0^1\int_0^1\, \frac{x(1-x)y(1-y)}3\, {\cal T}'\big( \frac{x(1-x)y(1-y)}{3} \big)\,{\rm d}x{\rm d}y.$$