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![[Post New]](/templates/default/images/icon_minipost_new.gif) 3 Mar 2007 21:33:38 IST
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Que.1 Find the sum of series : 1. 1+3/2+5/4+7/8+..........2n-1/2 n+......to  2. 1 + 1/(1+2) + 1/(1+2+3) + 1/(1+2+3+4) + ..........n terms Que.2 cos  + cos /sin 2 + cos /sin 4 + cos /sin 6 exists for what range of ? |
Umang |
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3 Mar 2007 22:17:43 IST
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2) 1 + 1/(1+2) + 1/(1+2+3) + tn=2/(n(n+1)) splitting the term 2 [ (n+1) - n] / [n(n+1)] = 2[ 1/n - 1/(n+1)] now put 1, 2 ,3 .............n and add terms would cancel out giving out 2[n-1] / n i am working on the other two i will post the answers tommorow
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Shashank nayak |
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3 Mar 2007 22:25:10 IST
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ignore this part of the above code now put 1, 2 ,3 .............n+1 and add terms would cancel out giving out 2n /( n+1)
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Shashank nayak |
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3 Mar 2007 23:18:16 IST
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i wil post it by tomorrow
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adversities cause some men to break other to break records............i m of the other type....... :-) |
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4 Mar 2007 00:38:18 IST
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Q1. Let x=1/2 Sum of n-terms is represented by Sn Sn= 1+3x+5x2+7x3+....................+(2n-1)xn-1 xSn= x+3x2+5x3+......................+(2n-3)xn-1 + (2n-1)xn Subtract xSn from Sn we get: (1-x)Sn=1+2x+2x2+2x3+................+2xn-1 - (2n-1)xn =1+ 2x/(1-x) [Since n tends to infinity, xn tends to zero as x<1] Sn=(1+x)/(1-x)2= 6 [Since x=1/2] Ans Q2. Solve as mentioned by Ankur
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4 Mar 2007 00:57:45 IST
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Que.2 cos + cos /sin 2 + cos /sin 4 + cos /sin 6 exists for what range of ? Observe the expression does not exists for sin 2=0 i.e. sin =0 i.e. = |
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4 Mar 2007 01:01:37 IST
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e.2 cos + cos /sin 2 + cos /sin 4 + cos /sin 6 exists for what range of ? Observe the expression does not exists for sin2=0 i.e. sin=0 i.e. = n where n=0,1,2,3 .... all integers Hence range of is all possible angles, exept integeral multiple of
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4 Mar 2007 12:45:29 IST
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1) Its an Arithmatico-geometric series and we have a standard result for it
for S= ab + (a+d)br+(a+2d)br2............+anbrn-1
S= ab + dbr(1-rn-1) - anbrn
------- ----------------- ---------
1-r (1-r)2 1-r
For ur Q, put
n=infinity
a=1
r=1/2
b=1
d=2
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5 Mar 2007 20:44:37 IST
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Hey all ! There is something missing in Que 2 . The corrected question is - The sum of series ,cos + cos /sin 2 + cos /sin 4 + cos /sin 6 + ......... exists for what range of ? The choices are - 1. 0< < 2. = 4. none I am sorry for the error !!!!!
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Umang |
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5 Mar 2007 21:09:30 IST
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Given summation is geometric.
cosQ[1/{1-(1/((sinQ)^2))}]= -tanQsinQ
Now both tanQ and sinQ should exist.So Q is not equal to pi/2.
So ans is 3).
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ADARSH
NITK Surathkal
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5 Mar 2007 21:12:43 IST
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Hey adarsh ! pls explain how u solved . Also , the answer given is (d)
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Umang |
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5 Mar 2007 21:16:29 IST
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Sorry the ans is 4). That is sum to infinity. Apply S=a/(1-r) Here r=1/(sinQ)^2 Now you get S= -tanQsinQ Both tanQ and sinQ should exist.So Q is not equal to pi/2. Did you type option 3) correctly???????
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ADARSH
NITK Surathkal
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5 Mar 2007 21:18:57 IST
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See , I am getting sum = -tanQsinQ so , for sum to exist , sin and tan , both should exist simultaneously , which is not possible . Hence , there is no value of Q . Am I right ?????
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Umang |
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5 Mar 2007 21:20:49 IST
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Oh ! I am sorry ! option 3 is pi/2 < Q < pi . I dont know what has happened to my typing ?????
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Umang |
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