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 $C _r={ }^{n} C _r$.
$(1989,5 M)$
Show Answer
Solution:
- $C _0-2^{2} \cdot C _1+3^{2} \cdot C _2-\ldots+(-1)^{n}(n+1)^{2} \cdot C _n$
$$ =\sum _{r=0}^{n}(-1)^{r}(r+1)^{2}{ }^{n} C _r=\sum _{r=0}^{n}(-1)^{r}\left(r^{2}+2 r+1\right)^{n} C _r $$
$$ \begin{aligned} & =\sum _{r=0}^{n}(-1)^{r} r^{2} \cdot{ }^{n} C _r+2 \sum _{r=0}^{n}(-1)^{r} r \cdot{ }^{n} C _r+\sum _{r=0}^{n}(-1)^{r} \cdot{ }^{n} C _r \\ & =\sum _{r=0}^{n}(-1)^{r} \cdot r(r-1) \cdot{ }^{n} C _r+3 \cdot \sum _{r=0}^{n}(-1)^{r} \cdot r \cdot{ }^{n} C _r{ } _n \end{aligned} $$
$$ +\sum _{r=0}^{n}(-1)^{r}{ }^{n} C _r $$
$=\sum _{r=2}^{n}(-1)^{r} n(n-1){ }^{n-2} C _{r-2}+3 \sum _{r=1}^{n}(-1)^{r} n \cdot{ }^{n-1} C _{r-1}$ $+\sum _{r=0}^{n}(-1)^{r}{ }^{n} C _r$
$=n(n-1){{ }^{n-2} C _0-{ }^{n-2} C _1+{ }^{n-2} C _2-\ldots+(-1)^{n}{ }^{n-2} C _{n-2} }$
$+3 n{-{ }^{n-1} C _0+{ }^{n-1} C _1-{ }^{n-1} C _2+\ldots+(-1)^{n}{ }^{n-1} C _{n-1} }$
$$ +{{ }^{n} C _0-{ }^{n} C _1+{ }^{n} C _2+\ldots+(-1)^{n}{ }^{n} C _n } $$
$=n(n-1) \cdot 0+3 n \cdot 0+0, \forall n>2=0, \forall n>2$
