00:01
So first one, we want to prove 1 over, we want to prove this equal to am plus 1, am plus 2 minus am plus 2, am plus 2, am plus 3.
00:19
So the right hand side is actually am plus 2 times am plus 1 minus am plus 3.
00:30
So this equal to am plus 2 times am plus 1 times am plus 3, am plus 3 minus am plus 1.
00:43
So this equal to, since we know from the condition that am plus 2 equal to am plus am plus am plus 1, this means am plus 3 equal to am plus 1, this means am plus 3 equal to am plus 1, plus am plus 2.
01:03
So this turns is equal to am plus 2, then this cancels with this turn.
01:11
This means it's one over aam plus 1, aam plus 3, equal to left -hand side.
01:21
So for the second one, we want to compute dimension and equal to zero to infinity, then aem plus 1, aem plus 3, so use the first sub -proper, we know it is n equal to 0 to infinity, am plus 1, am plus 2, minus, am plus 2, am plus 2, am plus 3.
01:50
So this will equal to, like summation, n equal to 0 to infinity, am plus 1, am plus 2, minus summation, am plus 1, am plus 2, minus summation, aemm, am plus 0 infinity, a .m...