If `a=1+2+4+…` up to `n` terms
`b=1+3+9+…` up to `n` terms
and `c=1+5+25+….` up to `n` terms.
then `Delta=|{:(a,2b,4c),(2,2,2),(2^(n),3^(n),5^(n)):}|=`

Options

A. `30^(n)`
B. `10^(n)`
C. 0
D. `2^(n) + 3^(n) + 5^(n)`

Answer

C

Explanation

We have,
`{:(a = 1 + 2 + 4,..."to n terms",= 2^(n) -1),(b = 1+ 3 + 9 +,..."to n terms",= (1)/(2) (3^(n) -1)),(c =1 + 5 + 25 +,..."to n terms",= (1)/(4) (5^(n) -1)):}`
`:. |(a,2b,4c),(2,2,2),(2^(n),3^(n),5^(n))|`
`= |(2^(n) -1,3^(n) -1,5^(n) -1),(2,2,2),(2^(n),3^(n),5^(n))|`
`= 2 |(2^(n) -1,3^(n) -1,5^(n) -1),(1,1,1),(1,1,1)|` Applying `R_(3) rarr R_(3) - R_(1)` and taking 2 common `R_(2)
`=2 xx 0 = 0`