2 votes 2 votes A function $f(x) $ satisfies $f(1)=3600$ and $f(1)+f(2)+ \dots + f(n) = n^2 f(n)$, for all positive integers $n>1.$ What is the value of $f(9)?$ $80$ $240$ $200$ $100$ $120$ Quantitative Aptitude cat2007 quantitative-aptitude functions + – go_editor asked Dec 7, 2015 • edited Mar 28, 2022 by Lakshman Bhaiya go_editor 13.8k points 2.8k views answer comment Share See all 0 reply Please log in or register to add a comment.
Best answer 2 votes 2 votes f1+f2=4f2 f2=f1/3 f1+f2+f3=9f3 f1+f1/3+f3=9f3 f3=f1/6 f1+f2+f3+f4=16f4 We get f4=f1/10 So we can say fn=f1/sum of n terms therefore f9=3600/45=80 Pooja Palod answered Dec 7, 2015 • edited Jan 13, 2017 by Vijay Thakur Pooja Palod 1.8k points comment Share See all 2 Comments See all 2 2 Comments reply Himanshu1 126 points commented Dec 8, 2015 reply Share @pooja : Answer should be 80 & that should be : fn=f1/sum.of n terms. 1 votes 1 votes tiger 54 points commented Dec 12, 2015 reply Share f9 = f1/45 = 3600/45 = 80 0 votes 0 votes Please log in or register to add a comment.