How to prove the following?

Question

(#n/k#) = (#(n-1)/(k-1)#) + (#(n-1)/(k)#)
Note that the bars are only there to separate the top and bottom. It's not a fraction (Pascals triangle)

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Answer 1 · 2017-12-08T07:15:07

Please refer to a Proof given in the Explanation.

Recall the Definition : #((n),(k))=(n!)/{(n-k)!k!}#.

#:." The R.H.S.="((n-1),(k-1))+((n-1),(k)),#

#=((n-1)!)/{((n-1)-(k-1))!(k-1)!}+((n-1)!)/{((n-1)-k)!k!},#

#=((n-1)!)/{(n-k)!(k-1)!} + ((n-1)!)/{(n-k-1)!k!}.........................(ast).#

Note that, #k! = 1.2.3.....(k-1).k=(k-1)!k............(ast_1),#

# rArr (n-k)! =(n-k-1)!(n-k)..............................(ast_2).#

Using #(ast_1) and (ast_2)" in "(ast),# we have,

#:." The R.H.S."=(n-1)![1/{(n-k-1)!(k-1)!(n-k)}+1/{(n-k-1)!(k-1)!k}],#

#=((n-1)!)/{(n-k-1)!(k-1)!}{1/(n-k)+1/k}.#

#=((n-1)!)/{(n-k-1)!(k-1)!}{((n-k)+k)/((n-k)k)},#

#=((n-1)!)/{(n-k-1)!(k-1)!}{n/((n-k)k)},#

#={(n-1)!n}/[{(n-k-1)!(n-k)}{(k-1)!k}],#

#=(n!)/[{(n-k)!}{k!}]............[because, (ast_1) and (ast_2)],#

#=((n),(k)).................................[because," the Definition],"#

#"=The L.H.S."#

Enjoy Maths.!