Find the value of `sum_(p=1)^(n) (underset(m=p)overset(n)sum""^(n)C_(m)""^(m)C_(p))`. And hence, find the value of `lim_(nrarroo) (1)/(3^(n)) sum_(m=p

Correct Answer - `(3^(n) - 2^(n))` and `1`
`S = underset(p=1)overset(n)sum(underset(m=p)overset(n)sum.^(n)C_(m).^(m)C_(p))`
`= underset(p=1)overset(n)sum(.^(n)C_(p).^(p)C_(p)+.^(n)C_(p+1).^(p)+1)C_(p) + "....."+.^(n)C_(n).^(n)C_(p))`
`= underset(p=1)overset(n)sum` [coefficient of `x^(p)` in `{.^(n)C_(p)(1+x)^(p)+.^(n)C_(p+1)(1+x)^(p+1)+"....."+.^(n)C_(n)(1+x)^(n)}]`
`= underset(p=1)overset(n)sum` [Coefficeint of `x^(p)` in `{.^(n)C_(0)+.^(n)C_(1)(1+x)+.^(n)C_(2)(1+x)^(2)+"....."+.^(n)C_(p-1)(1+x)^(p-1)+.^(n)C_(p)(1+x)^(p)+"....."+.^(n)C_(n)(1+x)^(n)}]`
`= underset(p=1)oversetg(n)sum` [coefficient of `x^(p)` in `{1+(1+x)}^(n)`]
`= underset(p=1)overset(n)sum` [coefficient of `x^(p)` in `(2+x)^(n)`]
`= underset(p=1)overset(n)sum[.^(n)C_(p)2^(n-p)]`
` = .^(n)C_(1)2^(n-1)+.^(n)C_(2)2^(n-2)+"...".^(n)C_(n)`
`= .^(n)C_(0)2^(n)+.^(n)C_(1)2^(n-1)+.^(n)C_(2)2^(n-2)+"...."+.^(n)C_(n)-.^(n)C_(0)2^(n)`
`= (1+2)^(n)-2^(n)`
`= 3^(n) - 2^(n)`
Alternate solution :
`underset(p=1)overset(n)sum(underset(m=p)overset(n)sum.^(n)C_(m).^(m)C_(p))= underset(p=1)overset(n)sum(underset(m=p)overset(n)sum(n!)/((n-m)!m!)(m!)/((m-p)!p!))`
`= underset(p=1)overset(n)sum(underset(m=p)overset(n)sum(n!)/((n-m)!)(1)/((m-p)!p!))`
`=underset(p=1)overset(n)sum(n!)/((n-p)!p!)(underset(m=p)overset(n)sum((n-p)!)/((n-m)!(m-p)!))`
`= underset(p=1)overset(n)sum.^(n)C_(P)(underset(m=p)overset(n)sum.^(n-p)C_(m-p))`
`= underset(p=1)overset(n)sum .^(n)C_(p)2^(n-p)`
`= .^(n)C_(1)2^(n-1)+.^(n)C_(2)2^(n-2)+"....."+.^(n)C_(n)2^(0)`
`= (2+1)^(n)-.^(n)C_(0)2^(n)`
`= 3^(n) - 2^(n)`
Also,
`underset(nrarroo)"lim"1/(3^(n)) underset(p=1)overset(n)sum(overset(n)underset(m=p)sum.^(n)C_(m).^(m)C_(p))= underset(nrarroo)"lim"(3^(n)-2^(n))/(3^(n)) = 1`