r4-3r+1=(r2-r-1)(r2+r-1)
tr=1/2(2r/r4-3r+1)
=1/2({(r2+r-1)/r4-3r+1}-{(r2-r-1)/r4-3r+1})
Tr=1/2[(1/r2-r-1)-1/r2+r-1]
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n thus applting summation ans=1/2(-1)
[1]
Let rth termof the series be given by Tr= r/1-3r2+r4 then \lim_{n\rightarrow \propto \sum_{r=1}^{n}{T_{r}}} is :-
a) 3/2
b) 1/2
c) -1/2
d) -3/2
ANS :- (c)
PLEASE POST THE SOLUTION IF SOLVED....
r4-3r+1=(r2-r-1)(r2+r-1)
tr=1/2(2r/r4-3r+1)
=1/2({(r2+r-1)/r4-3r+1}-{(r2-r-1)/r4-3r+1})
Tr=1/2[(1/r2-r-1)-1/r2+r-1]
..
.
.
.
.
.
n thus applting summation ans=1/2(-1)
[1]