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Prove that 1/(m!).^(n)C(0)+(n)/((m+1)!...

Prove that
`1/(m!).^(n)C_(0)+(n)/((m+1)!).^(n)C_(1)+(n(n-1))/((m+2)!).^(n)C_(2)+"....."+(n(n-1)"...."2xx1)/((m+2)!).^(n)C_(n) = ((m+n+1)(m+n+2)"....."(m+2n))/((m+n)!)`

A

1

B

2

C

3

D

4

Text Solution

Verified by Experts

The correct Answer is:
a
We have,
`""^(m)C_(r)""^(n)C_(0)+""^(m)C_(r-1)""^(n)C_(1) + ""^(m)C_(r-2) ""^(n)C_(2) +...+ ""^(m)C_(0)""^(n)C_(r) = ""^(m+n)C_(r)`
= Coefficient of `x^(r)` in `{ (1 + x)^(m) (1 + x)^(n)}`
= Coefficient of `x^(r)` in ` (1 + x)^(m + n) = ""^(m+n)C_(r)`
So, statement-2 is true.
Now,
`(1)/(m!) C_(0) + (n)/((m+1)!) C_(1) (n(n-1))/((m+2)!) C_(2) +... + (n(n-1) (n -2)...2.1)/((m +n)!) C_(n)`
`= (n!)/((m + n)!) { ((m+n)!)/(m!n!) ""^(n)C_(0) + ((m +n)!)/((m+1)!( n-1)!) ""^(n)C_(1)`
` + ((m + n)!)/((n-2)!) ""^(r)C_(2) +...+ ((m+n)!)/((m+n)!) ""^(n)C_(n)}`
`= (n!)/((m+n)!) {""^(m+n)C_(n-2) ""^(n)C_(0) + ""^(m+n)C_(n-1) ""^(n)C_(1) + ""^(m+n)C_(n-2)""^(n)C_(2) +...+""^(m+n)C_(0)""^(n)C_(n)}`
`= (n!)/((m+n)!)""^(m+n+n)C_(n)` [Using statement-2]
`= (n!)/((m+n)!) xx((m + 2n)!)/((m+n)!n!) `
`=((m +n+1)(m +n+2)(m+ 2n))/((m+n)!) `
So, statement-1 is also true. Statement-2 is a correct
expanation for statement-1.
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