Binomial Theorem 2 Question 14
17.
Prove that $C_{0}-2^{2} \cdot C_{1}+3^{2} \cdot C_{2}-\ldots+(-1)^{n}(n+1)^{2} \cdot C_{n}$ $=0, n>2$, where $C_{r}={ }^{n} C_{r}$.
$(1989,5 \mathrm{M})$
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Solution:
- $C_{0}-2^{2} \cdot C_{1}+3^{2} \cdot C_{2}-\ldots+(-1)^{n}(n+1)^{2} \cdot C_{n}$
$ =\sum_{r=0}^{n}(-1)^{r}(r+1)^{2}{ }^{n} C_{r}=\sum_{r=0}^{n}(-1)^{r}\left(r^{2}+2 r+1\right)^{n} C_{r} $
$ =\sum_{r=0}^{n}(-1)^{r} r^{2} \cdot{ }^{n} C_{r}+2 \sum_{r=0}^{n}(-1)^{r} r \cdot{ }^{n} C_{r}+\sum_{r=0}^{n}(-1)^{r} \cdot{ }^{n} C_{r} $
$=\sum_{r=0}^{n}(-1)^{r} \cdot r(r-1) \cdot{ }^{n} C_{r}+3 \cdot \sum_{r=0}^{n}(-1)^{r} \cdot r \cdot{ }^{n} C_{r}{ }_{n}$
$ +\sum_{r=0}^{n}(-1)^{r}{ }^{n} C_{r} $
$=\sum_{r=2}^{n}(-1)^{r} n(n-1){ }^{n-2} C_{r-2}+3 \sum_{r=1}^{n}(-1)^{r} n \cdot{ }^{n-1} C_{r-1}$ $+\sum_{r=0}^{n}(-1)^{r}{ }^{n} C_{r}$
$=n(n-1){{ }^{n-2} C_{0}-{ }^{n-2} C_{1}+{ }^{n-2} C_{2}-\ldots+(-1)^{n}{ }^{n-2} C_{n-2}}$
$+3 n{-{ }^{n-1} C_{0}+{ }^{n-1} C_{1}-{ }^{n-1} C_{2}+\ldots+(-1)^{n} \quad { }^{n-1} C_{n-1}}$
$ +{{ }^{n} C_{0}-{ }^{n} C_{1}+{ }^{n} C_{2}+\ldots+(-1)^{n} \quad{ }^{n} C_{n}} $
$=n(n-1) \cdot 0+3 n \cdot 0+0, \forall n>2=0, \forall n>2$
