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(n-1)Pr+r.n-1)P(n-1) is equal to...

`(n-1)P_r+r.n-1)P_(n-1)` is equal to

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Prove that : ^nP_r= "^(n-1)P_r+r ^(n-1)P_(r-1) , for all natural numbers n and r for which the symbols are defined.

Show that ""^(n)P_r =""^(n-1) P_r + r, ^(n-1) P_(r-1) Where the symbols have their usual meanings.

Prove that ^nP_r= ^(n-1)P_r+r^(n-1)P_(r-1) (notation used are in their usual meaning).

Prove that .^(n)P_(r)=.^(n-1)P_(r)+r.^(n-1)P_(r-1) .

What is ""^(n-1)P_(r )+ ""^(n-1)P_(r-1) ?

Prove that .^(n-1)P_r+r.^(n-1)P_(r-1)=^nP_r .

Prove that .^(n-1) P_r+r .^(n-1) P_(r-1) = .^nP_r

Prove that ^(n-1) P_r+r .^(n-1) P_(r-1) = .^nP_r

The value of sum_(r=1)^n(sum_(p=0)^(r-1) ^nC_r ^rC_p 2^p) is equal to (a) 4^(n)-3^(n)+1 (b) 4^(n)-3^(n)-1 (c) 4^(n)-3^(n)+2 (d) 4^(n)-3^(n)