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  • If <img alt="(1+x)^{n}=C_{0}+C_{1} x+C_{2} x^{2}+C_{3} x^{3}+\ldots+C_{n} x^{n}" src="http://entrancecorner.oncodecogs.com/gif.latex?%281&plus;x%29%5E%7Bn%7D%3DC_%7B0%7D&plus;C_%7B1%7D

If (1+x)^{n}=C_{0}+C_{1} x+C_{2} x^{2}+C_{3} x^{3}+\ldots+C_{n} x^{n} then, \mathrm{C}_{0} \mathrm{C}_{\mathrm{r}}+\mathrm{C}_{1} \mathrm{C}_{\mathrm{r}+1}+\mathrm{C}_{2} \mathrm{C}_{\mathrm{r}+2}+\ldots+C_{n-\mathrm{r}} \mathrm{C}_{n} is equal to?

Option: 1

\frac{2 n !}{(n+r) !(n+r) !}


Option: 2

\frac{2 n !}{(n-r) !(n-r) !}


Option: 3

\frac{2 n !}{(n) !(r) !}


Option: 4

\frac{2 n !}{(n-r) !(n+r) !}


Answers (1)

best_answer

\mathrm{C}_{0} \mathrm{C}_{\mathrm{r}}+\mathrm{C}_{1} \mathrm{C}_{\mathrm{r}+1}+\mathrm{C}_{2} \mathrm{C}_{\mathrm{r}+2}+\ldots+C_{n-\mathrm{r}} \mathrm{C}_{n}

=\mathrm{C}_{0} \mathrm{C}_{\mathrm{n-r}}+\mathrm{C}_{1} \mathrm{C}_{\mathrm{n-r-1}}+\mathrm{C}_{2} \mathrm{C}_{\mathrm{n-r-2}}+\ldots+C_{n-\mathrm{r}} \mathrm{C}_{0}

= ^{n+n}C_{n-r} = ^{2n} C_{n-r}

This equals option D

Posted by

shivangi.bhatnagar

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