During my resarch, i checked using mapple that we have numericall for $ 2 \leq n \leq 20,$ $ \forall t $ integer satisfying $ 0 \leq t \leq n-1$

$$ \displaystyle \frac{1}{2} \displaystyle \sum_{p=1}^{n-t} C_{n}^{p+t} C_{n+p+t}^{n} \displaystyle \frac{(-1)^{p}}{p}= \displaystyle C_{n+t}^n\displaystyle \sum_{p=1}^{n-t} C_{n}^{p+t} \displaystyle \frac{(-1)^{p}}{p}$$

i think that this true for all integer $n \geq 2$ and forall $t$ integer satisfying $\forall 0 \leq t \leq n-1$ and i need a proof... any help?

is it known result? ps: here $ \forall m,k$ integer with $0 \leq k \leq m, C_m^k=\displaystyle \frac{m!}{k!(m-k)!}$

During my research, I checked using Maple that we have numerically for $ 2 \leq n \leq 20,$ $ \forall t $ integer satisfying $ 0 \leq t \leq n-1$

$$ \displaystyle \frac{1}{2} \displaystyle \sum_{p=1}^{n-t} C_{n}^{p+t} C_{n+p+t}^{n} \displaystyle \frac{(-1)^{p}}{p}= \displaystyle C_{n+t}^n\displaystyle \sum_{p=1}^{n-t} C_{n}^{p+t} \displaystyle \frac{(-1)^{p}}{p}$$

I think that this true for all integer $n \geq 2$ and forall $t$ integer satisfying $\forall 0 \leq t \leq n-1$ and i need a proof... any help?

Is it known result? ps: here $ \forall m,k$ integer with $0 \leq k \leq m, C_m^k=\displaystyle \frac{m!}{k!(m-k)!}$

During my resarch, i checked using mapple that we have numericall for $ 2 \leq n \leq 20,$ $ \forall t $ integer satisfying $ 0 \leq t \leq n-1$

$$ \displaystyle -\frac{1}{2} \displaystyle \sum_{p=1}^{n-t} C_{n}^{p+t} C_{n+p+t}^{n} \displaystyle \frac{(-1)^{p+t}}{p}= \displaystyle (-1)^t C_{n+t}^n\displaystyle \sum_{p=1}^{n-t} C_{n}^{p+t} \displaystyle \frac{(-1)^{p+1}}{p}$$

i think that this true for all integer $n \geq 2$ and forall $t$ integer satisfying $\forall 0 \leq t \leq n-1$ and i need a proof... any help?

is it known result? ps: here $ \forall m,k$ integer with $0 \leq k \leq m, C_m^k=\displaystyle \frac{m!}{k!(m-k)!}$

During my resarch, i checked using mapple that we have numericall for $ 2 \leq n \leq 20,$ $ \forall t $ integer satisfying $ 0 \leq t \leq n-1$

$$ \displaystyle \frac{1}{2} \displaystyle \sum_{p=1}^{n-t} C_{n}^{p+t} C_{n+p+t}^{n} \displaystyle \frac{(-1)^{p}}{p}= \displaystyle C_{n+t}^n\displaystyle \sum_{p=1}^{n-t} C_{n}^{p+t} \displaystyle \frac{(-1)^{p}}{p}$$

i think that this true for all integer $n \geq 2$ and forall $t$ integer satisfying $\forall 0 \leq t \leq n-1$ and i need a proof... any help?

is it known result? ps: here $ \forall m,k$ integer with $0 \leq k \leq m, C_m^k=\displaystyle \frac{m!}{k!(m-k)!}$