Prove that $^{n} C_{0} -^{n} C_{1} +^{n} C_{2} -^{n} C_{3} + . . . + {(- 1)}^{r}^{n} C_{r} + . . . = {(- 1)}^{r - 1}^{n - 1} C_{r - 1}$

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Solution

The statement given to prove is wrong as the L.H.S of the question evaluates to zero on solving. It does not depend on r.

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Similar questions

Let $S_{1}$ = $^{n} C_{0}$ + $^{n} C_{1}$ + $^{n} C_{2}$ ............. $^{n} C_{n}$ and $S_{2}$ = $^{n} C_{0}$ - $^{n} C_{1}$ + $^{n} C_{2}$ ..............+ $(- 1)^{n}$ $^{n} C_{n}$

Find the value of $\frac{S_{1}}{S_{1} + S_{2}}$ is ___.

Q. If

(1 + x)^{n} =^{n} C_{0} +^{n} C_{1} x +^{n} C_{2} x^{2} +^{n} C_{n} x^{n}, x ϵ N

then prove that

^{n} C_{0} -^{n} C_{1} +^{n} C_{2} - . . . . . . + (- 1)^{m - 1} \times^{n} C_{m - 1}

= (- 1)^{m - 1} \frac{(n - 1) (n - 2) . . (n - m + 1)}{(m - 1)!}

If $S_{1}$ = $^{n} C_{0}$ + $^{n} C_{1}$ + $^{n} C_{2}$ .................. $^{n} C_{n}$ and $S_{2}$ = $^{n} C_{0}$ - $^{n} C_{1}$ + $^{n} C_{2}$ .................... $^{n} C_{n}$