JEE corner (mains + advanced) #1
f ( 2 0 0 9 1 ) + f ( 2 0 0 9 2 ) + f ( 2 0 0 9 3 ) + … + f ( 2 0 0 9 2 0 0 8 )
If f ( x ) = 1 + e 2 x − 1 e 2 x − 1 , then find the value of the above expression.
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
5 solutions
Aakash lol
Log in to reply
hmm @Eashan Mundhe i go to aakash so i felt lazy while writing the ans. so i uploaded the screenshot.... :P
Log in to reply
but aakash institute is not good for IIT preparation. please leave it
f ( 2 0 0 9 1 ) + f ( 2 0 0 9 2 0 0 8 ) = f ( 2 0 0 9 2 ) + f ( 2 0 0 9 2 0 0 7 ) = f ( 2 0 0 9 3 ) + f ( 2 0 0 9 2 0 0 6 ) = . . . f ( 2 0 0 9 1 0 0 4 ) + f ( 2 0 0 9 1 0 0 5 ) = 1
Thus, the sum is 1004.
By using the law of Gauss, Sn=n/2(a+l) n=count of term a=first term l=last term Here, n=2008 a=f(1/2009) l=f(2008/2009) So S = 2008/2(f(1/2009)+f(2008/2009)) = 1004(1) = 1004
how can we prove that it is an AP?
Let..e^2x-1=p Whatever f(x)=whatever its no matter So..the equation will be p/(1+p) And the sum will be 2008p/(1+p) Let p=1 So...2008×1/(1+1)=1004
Take the average of the first and last terms and then multiply the result by 2008 since there are 2008 terms