First of all, let us observe that the sum is nonzero only for , and , i.e. only for such k for which and . Such k exists only for and , i.e. for and . Hence, if we investigate the term , we find out that
(*) , where is the Kronecker delta.
Let us now proceed to the induction and consider . Then
,
since only and are allowed.
Further, let us assume, that the investigated equality is fullfilled for and we show that it holds for , as well. Thus, we start with
.
Now we apply the inductive step and write
.
The last sum vanishes for , hence we may consider only . In addition, if we take into account, that
, then there holds
.
Since for the last sum it holds
,
we obtain
and employing the equality (*) completes the proof.
for all 
and prove it by induction.
, 
and 
, i.e. only for such k for which 
and 
. Such k exists only for 
and 
, i.e. for 
and 
. Hence, if we investigate the term 
, we find out that
, where 
is the Kronecker delta.
. Then 
,
and 
are allowed. 
and we show that it holds for 
, as well. Thus, we start with![$\binom{n}{p}\binom{n}{q} = \left[\binom{n-1}{p}+\binom{n-1}{p-1}\right]\left[\binom{n-1}{q}+\binom{n-1}{q-1}\right] = $](/mathtex/e6/e6c8d8b7932cd593023450baf231482e.gif "kopírovat do textarea")
.
![$=\sum^{n-1}_{k=0}\binom{p}{k}\binom{q}{k}\left[\binom{n+k}{p+q}-{\color{red}\binom{n-1+k}{p+q-1}} \right] + \sum^{n-1}_{k=0}\binom{p-1}{k}\binom{q}{k}{\color{red}\binom{n-1+k}{p+q-1}} +$](/mathtex/ad/ad12c8980dc8284f68ff5f5a724a2bb5.gif "kopírovat do textarea")
![$+ \sum^{n-1}_{k=0}\binom{p}{k}\binom{q-1}{k}{\color{red}\binom{n-1+k}{p+q-1}}+ \sum^{n-1}_{k=0}\binom{p-1}{k}\binom{q-1}{k}\left[\binom{n+k}{p+q-1}-{\color{red}\binom{n-1+k}{p+q-1}}\right] =$](/mathtex/26/26718bac2a01939f1d343601795831e7.gif "kopírovat do textarea")
![$+ \sum^{n-1}_{k=0} \left[-\binom{p}{k}\binom{q}{k}+\binom{p-1}{k}\binom{q}{k}+\binom{p}{k}\binom{q-1}{k}-\binom{p-1}{k}\binom{q-1}{k}\right]{\color{red}\binom{n-1+k}{p+q-1}}$](/mathtex/8b/8b7ac49771b6b7c0ed876e815a613ed1.gif "kopírovat do textarea")
.
, hence we may consider only 
. In addition, if we take into account, that
![$ = -\left[\binom{q}{k}-\binom{q-1}{k}\right]\left[\binom{p}{k}-\binom{p-1}{k}\right] = -\binom{q-1}{k-1}\binom{p-1}{k-1}$](/mathtex/44/4427809d32416c43ad4ff8c08fb8939a.gif "kopírovat do textarea")
, then there holds
.
,