` sum_(r=1)^(n) {sum_(r1=0)^(r-1) ""^(n)C_(r) ""^(r)C_(r_(1)) 2^(r1)}` is equal to

Options

A. `4^(n) -3^(n) +1`
B. `4^(n) - 3^(n) -1`
C. `4^(n) - 3^(n) +2`
D. `4^(n) - 3^(n)`

Answer

d

Explanation

We have,
` sum_(r=1)^(n) (sum_(r1=0)^(r-1) ""^(n)C_(r) ""^(n)C_(r_(1)) 2^(r)1)`
=` sum_(r=1)^(n) ""^(n)C_(r) (sum_(r1=0)^(r-1) ""^(n)C_(r_(1)) 2^(r)1)`
=` sum_(r=1)^(n) ""^(n)C_(r) {( 1 + 2 )^(r) - 2^(r)}`
=` sum_(r=1)^(n) ""^(n)C_(r) 3^(r) -sum_(r=1)^(n) ""^(n)C_(r) 2^(r)`
` {(1 + 3)^(n) -1} - {( 1 + 2)^(n) - 1} = 4^(n) - 3^(n)`.