For a positive integer n,
let `a(n)=1+(1)/(2)+(1)/(3)+(1)/(4)+ . . .+(1)/(2^(n)-1)` Then:

Options

A. `a(100)lt100`
B. `a(100)gt100`
C. `a(200)gt100`
D. `a(200)lt100`

Answer

A, C

Explanation

`:.a(n)=1+(1)/(2)+(1)/(3)+(1)/(4)"......"+(1)/(2^(n)-1)`
`=1+((1)/(2)+(1)/(3))+((1)/(4)+(1)/(5)+(1)/(6)+(1)/(7))+((1)/(8)+"......"+(1)/(15))+"......"+(1)/(2^(n)-1)`
`=1+((1)/(2)+(1)/(2^(2)-1))+((1)/2^(2)+(1)/(5)+(1)/(6)+(1)/(2^(3)-1))+"......"+(1)/(2^(n)-1)`
`:. a(n)lt1+1+"....."+` n terms
`implies a(n)ltn`
`implies a(100)lt100`
Also, `a(n)=1+(1)/(2)+((1)/(3)+(1)/(4))+((1)/(5)+(1)/(6)+(1)/(7)+(1)/(8))+"......"+(1)/(2^(n)-1)`
`=1+(1)/(2)+((1)/(2^(1)+1)+(1)/(2^(2)))+((1)/(2^(2)+1)+(1)/(6)+(1)/(7)+(1)/(2^(3)))+"......"+((1)/(2^(n-1)+1)+"......"+(1)/(2^(n)))-(1)/(2^(n))`
`a(n)gt1+(1)/(2)+(2)/(4)+(4)/(8)+"......."+(2^(n-1))/(2^(n))-(1)/(2^(n))`
`a(n)gt(1-(1)/(2^(n)))+(n)/(2)" " implies a(n)gt(n)/(2)`
`:.a(200)gt100`