Dear student There is a mistake in the question ∑r=0n(-1)r nCr 2n-r xrWe know ∑r=0n-1r nCran-rbr=(a-b)nOn Comparing, we get2n-r=an-r and xr= br⇒a=2 and b=xSo, ∑r=0n(-1)r nCr 2n-r xr=(2-x)n Regards

sir
sir 2-1) Choose one: O O O O 012 (x-2)'1 n c H is equivalent to

Dear student

There is a mistake in the question . \sum_{r = 0}^{n} (- 1)^{r}^{n} C_{r} 2^{n - r} x^{r} We know \sum_{r = 0}^{n} {(- 1)}^{r}^{n} C_{r} a^{n - r} b^{r} = (a - b)^{n} On Comparing, we get 2^{n - r} = a^{n - r} and x^{r} = b^{r} \Rightarrow a = 2 and b = x So, \sum_{r = 0}^{n} (- 1)^{r}^{n} C_{r} 2^{n - r} x^{r} = (2 - x)^{n}

Regards

View Full Answer

​sir​sir 2-1) Choose one: O O O O 012 (x-2)'1 n c H is equivalent to

sir
sir 2-1) Choose one: O O O O 012 (x-2)'1 n c H is equivalent to