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Binomial Theorem question (1 Viewer)
- Thread starter apollo1
- Start date
someth1ng
Retired Nov '14
Just out of curiousity, what question was this in? (1-7)
we know that:
^r(_{k}^{n}\textrm{C})(_{2r-k}^{n}\textrm{C})= (-1)^r(_{r}^{n}\textrm{C}))
from i)
Now we break the above sum on the LHS to:
^r(_{k}^{n}\textrm{C})(_{2r-k}^{n}\textrm{C})= \sum_{k=0}^{r}(-1)^r(_{k}^{n}\textrm{C})(_{2r-k}^{n}\textrm{C}) + \sum_{r}^{2r}(-1)^r(_{k}^{n}\textrm{C})(_{2r-k}^{n}\textrm{C}) - (-1)^r(_{r}^{n}\textrm{C})^2)
realising that there are two terms for the rth term so we minus by the rth term which is
^r(_{r}^{n}\textrm{C})^2)
now we realise if we expand the now broken sum we can deduce that:
^r(_{k}^{n}\textrm{C})(_{2r-k}^{n}\textrm{C}) = \sum_{r}^{2r}(-1)^r(_{k}^{n}\textrm{C})(_{2r-k}^{n}\textrm{C}))
as there are an equal number of terms in each sum and by symmetry of combinations we will get the same values.
Therefore:
^r(_{k}^{n}\textrm{C})(_{2r-k}^{n}\textrm{C}) - (-1)^r(_{r}^{n}\textrm{C})^2= (-1)^r(_{r}^{n}\textrm{C}))
and if we rearrange this and factorise we obtain:
^r(_{k}^{n}\textrm{C})(_{2r-k}^{n}\textrm{C}) = (1/2)(-1)^r(_{r}^{n}\textrm{C})[1 + (_{r}^{n}\textrm{C})])
which is the desired result
from i)
Now we break the above sum on the LHS to:
realising that there are two terms for the rth term so we minus by the rth term which is
now we realise if we expand the now broken sum we can deduce that:
as there are an equal number of terms in each sum and by symmetry of combinations we will get the same values.
Therefore:
and if we rearrange this and factorise we obtain:
which is the desired result