Hi.
Skrytý text:We claim that
for all
and prove it by induction.
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.