1.)
limn→∞ [1/n+ n2/(n+1)3+ n2/n+23.... 1/8n ]
2.)
lim 1/n[sec2π/4n+ sec22π/4n...sec2nπ/4n]
n→∞
3.)
limn→∞
61st one is easy
here is the solution:
limn→∞ [1/n+ n2/(n+1)3+ n2/n+23.... 1/8n ]
am omitting limit but it is there untill i say ...
=1/n[n/n+n3/n+13+.......1/8]
this can be written as
1/n [n3/n3+n3/n+13+.....n3/8n3]
which equals
lim n→∞ Σr=0 to n 1/(1+r/n)3
let x=r/n
therefore dx=1/n
S=∫0 to 1 dx/(1+x)3
=1+x-3+1/-3+1 from 0 to 1
Putting limits we get answer as 3/8....
hope solution is okay!!
11>the first term caN BE WRITTEN AS
n2(n+0)3 and then integ
n2(n+r)3 from 0 to 1
1is ans of 1st 3/8????using the 1st principle of integration....
and ans of 2nd 4/pie????Again using the 1st principle of integration....
62nd one same way yarr!! its easy....
lim n→∞ Σ r=1 to n (sec rπ/4n)2
taking 1/n=dx
we get
S= ∫0 to 1 (sec πx /4)2
which comes out to be
4/π[tan π/4x]0 to 1
therefore S= 4/Ï€
This should be the answer!!!
6yeah @lanuk iam getting same answer as that of urs!!!