For a positive integer `n` let `a(n)=1+1/2+1/3+1/4+1/((2^n)-1)dot` Then `a(100)lt=100` b. `a(100)dot> 100` c. `a(200)lt=100` d. `a(200)lt=100`

Options

A. `a (100) le 100`
B. `a (100) gt 100`
C. `a (200) le 100`
D. `a (200) gt 100`

Answer

A, D

Explanation

Given, `a (n) = 1 + (1)/(2) + (1)/(3) + (1)/(4) + ...+ (1)/(2^(n) -1)`
`= 1 + ((1)/(2) + (1)/(3)) + ((1)/(4) + ...+ (1)/(7)) + ((1)/(8) + ...+ (1)/(15)) + ...+ ((1)/(2^(n-1)) + ...+ (1)/(2^(n) -1))`
`lt 1 + ((1)/(2) + (1)/(2)) + ((1)/(4) + (1)/(4) + ...+ (1)/(4)) + ((1)/(8) + (1)/(8) + ...+ (1)/(8)) + ..+ ((1)/(2^(n-1)) + ....+ (1)/(2^(n-1)))`
`= 1 + (2)/(2) + (4)/(4) + (8)/(8) + ...+ (2^(n-1))/(2^(n-1))`
`= ubrace(1 + 1 + 1 + 1 + ...+ 1)_("(n) times") =n`
Thus, `a (100) le 100`
Therefore, (a) is the answer.
Again, `a(n) = 1 + (1)/(2) + ((1)/(3) + (1)/(4)) + ((1)/(5) + ...+ (1)/(8)) + ...+ ((1)/(2^(n-1) +1) + ...+ (1)/(2^(n))) - (1)/(2^(n))`
`gt1 + (1)/(2) + ((1)/(4) + (1)/(4)) + ((1)/(8) + (1)/(8) + ...+ (1)/(8)) + ..+ ((1)/(2^(n)) + ...+ (1)/(2^(n))) - (1)/(2^(n))`
`= 1 + (1)/(2) + (2)/(4) + (4)/(8) + ...+ (2^(n-1))/(2^(n)) - (1)/(2^(n))`
`= 1 + ubrace((1)/(2) + (1)/(2) + (1)/(2) + ...+ (1)/(2))_("n times") - (1)/(2^(n)) = (1 -(1)/(2^(n))) + (n)/(2)`
Therefore, `a (200) gt (1 -(1)/(2^(200))) + (200)/(2) gt 100`
Therefore, (d) is also the answer