33[12] ok...
Today's FULL Test XI AITS question.. [7]
1. Find..
lim (sin4x+sin4(2x)+...+sin4(2nx))
n→∞ 4 4n
-
UP 0 DOWN 0 2 12
12 Answers
62try adding and substracting
cos4x
and if thta does not work then add and substract
cos4(2nx)
4n
62arey even I am not sure.. this is the only thing I could think ;)
33dekhte hain...
iska soln diya hai.. par dekh kar maja nahi aaya.. kyunki waisa mere se kabhi nhai hoga xam me apne se...
1dude can u pl tell da question in words .... I'm not getting da question?????
1priyam .. i made this last expression u wrote...
then in vain ....
but again ... isnt this in GP ??
i got something... ticked none f these... got -ve... :(
341Nice problem, seeing one in trig after a long time
sin2 θ = (1- cos 2θ)/2
sin4θ = (1-cos 2θ)2/4 = 1/8 + cos 2θ/2 - cos 4θ/8)
Hence the given summation is r=0Σn 1/4r(1/8 + cos 2r+1θ/2 - cos 2r+2θ/8) = 1/8 r=0Σn 1/4r + r=0Σn 1/4r(cos 2r+1θ/2 - cos 2r+2θ/8)
r=0Σn 1/4r(cos 2r+1θ/2 - cos 2r+2θ/8) is a telecopic sum which sums to cos 2θ/2 - cos 2n+2θ / 2.4n+1
In the limit it becomes cos 2θ/2.
1/8 r=0Σn 1/4r is a GP which in the limit is 1/6
So, the limit is 1/6 + cos 2θ/2
341Oh! There is a mistake in my working. The expression for sin4θ should actually be
3/8 -cos 2θ/2 + cos 4θ/8
So the summation will work out, by the same procedure as above, to 1/2 - cos 2θ/2 = sin2θ
Thanks for pointing out (though I should have verified as the limit is 0 when x = 0)
