Question

2(D):16. Prove that :
"C, + "C,-1 = "+IC,​

Answer 1

Appropriate Question :-

Prove that,

$\sf \: ^{n}C_{r} \: + \: ^{n}C_{r - 1} \: = \: ^{n + 1}C_{r}$

$\green{\large\underline{\sf{Solution-}}}$

Consider LHS

$\rm :\longmapsto\: \sf \: ^{n}C_{r} \: + \: ^{n}C_{r - 1}$

$\rm \:  =  \: \dfrac{n!}{r! \: (n - r)!} + \dfrac{n!}{(r - 1)! \: (n - r + 1)!}$

$\rm \:  =  \: \dfrac{n!}{r(r - 1)! \: (n - r)!} + \dfrac{n!}{(r - 1)! \: (n - r + 1)(n - r)!}$

$\rm \:  =  \: \dfrac{n!}{(r - 1)! \: (n - r)!} \bigg[\dfrac{1}{r} + \dfrac{1}{n - r + 1} \bigg]$

$\rm \:  =  \: \dfrac{n!}{(r - 1)! \: (n - r)!} \bigg[\dfrac{n - r + 1 + r}{r(n - r + 1)} \bigg]$

$\rm \:  =  \: \dfrac{n!}{(r - 1)! \: (n - r)!} \bigg[\dfrac{n+ 1}{r(n - r + 1)} \bigg]$

$\rm \:  =  \: \dfrac{(n + 1)n!}{r(r - 1)! \: (n - r + 1)(n - r)!}$

$\rm \:  =  \: \dfrac{(n + 1)!}{r! \: (n - r + 1)!}$

$\rm \:  =  \: \dfrac{(n + 1)!}{r! \: (n + 1- r)!}$

$\rm \:  =  \: ^{n + 1}C_{r}$

Hence,

$\rm :\implies\:\red{ \boxed{ \sf{ \:\sf \: ^{n}C_{r} \: + \: ^{n}C_{r - 1} \: = \: ^{n + 1}C_{r}}}}$

Additional Information :-

$\red{ \boxed{ \sf{ \:^{n}C_{r} = \frac{n! }{r! \: (n - r)!} }}}$

$\red{ \boxed{ \sf{ \: \frac{^{n}C_{r}}{^{n}C_{r - 1}} = \frac{n - r + 1}{r}}}}$

$\red{ \boxed{ \sf{ \:^{n}C_{r} = \frac{n}{r} \: ^{n - 1}C_{r - 1}}}}$

$\red{ \boxed{ \sf{ \:^{n}C_{r} \: = \: ^{n}C_{n - r}}}}$