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If `f(x)=sum_(r=1)^(x) 1/r` and `sum_(r=1)^(2017)f(r)=af(b)+c` thenA. `b=c`B. `a-b=1`C. `b+c=0`D. `a+c=0` |
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Answer» Correct Answer - B::C `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` |
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