If `f(x)=sum_(r=1)^(x) 1/r` and `sum_(r=1)^(2017)f(r)=af(b)+c` then

Options

A. `b=c`
B. `a-b=1`
C. `b+c=0`
D. `a+c=0`

Answer

B, C

Explanation

`f(k)=1+1/2+1/3+….+1/k`
`f(1)=1`
`f(2)=1+1/2`
`f(2017)=1+1/2+1/3+………...1/2017` on adding
`=1xx2017+1/2xx2016+1/3xx2015+…….+1/2017xx1`
`=1xx2017+1/2(2017-1)+1/3(2017-2)+…………….+1/2017(2017-2016)`
`=1xX2017(1+1/2+1/3+……+1/2017)-(1/2+2/3+.....+2016/2017)`
`=2017f(2017)-{1- 1/2+1- 1/3++1-1/4+....+1-1/2017}`
`=2017f(2017)-1xx2016+(1+1/2+............+1/2017)`
`=2017f(2017)-2016+f(2017)`
`=2018f(2017)-2017`
`a=2018, b=2017, c=-2017`