1mainly guys like snachit ,champ, tushar and all the edudigm ones are producing this trend !!!
Ques1) Prove that the sum of 1 st n odd natural numbers is n2 and hence or otherwise prove that the sum of next 'n' odd natural numbers is divisible by 3.
Ques2) Show that the sum of
2/3 + 8/9 + 26/27 + 80/81 + .............upto n terms is given by
[n - (3 n-1)] / 2(3 n).
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10 Answers
341Sorry if I sound rude, but I think you should be attempting these problems on your own. You are not going to gain anything by taking our solutions and copying it down into some assignment sheet.
If you can show your working and we can help you on. I dont think anyone of us should post complete solutions for these.
66Yes, that's right. The trend which I have seen recently concerning the posts is a bit disturbing. The forum is becoming more of a put-your-homework-to-get-it-solved place.
1
62I echo the sentiments of Prophet sir and Kaymant sir.
I absolutely agree.
The discussion part is not here...
The reason i think has got to be the timing.. we are too far from JEE and most students hardly know much in class XI at tihs stage even thought the tutions at most places have taught 5-10 chapters in all the subjects...
even in XII if i remember my own time, ppl wake up after september...
and last yr this forum became very active towards January!
1Correct, only towards january nice discussions get started, but i still wonder why is it so. ?
1For the second one, I am posting my working.
2/3 + 8/9 + 26/27 + 80/81 + .................. upto n terms
See the denomianotor side, 3,9,27,81, ......... forms g.p with first term a=3, and common ratio r = 9/3 = 3.
So, the n th terms of the denominator side is given by a ( r n-1) / (r-1)
= [ 3 ( 3 n -1)] / (3-1)
= [ 3 ( 3 n -1)] / 2
= (3/2) [ 3 n -1]
But my problem is with the numerator side, bcoz neither it forms A.P, nor G.P, nor A.G.P, nor Method of diff can be appplied.
So , plzz help me out to get the n th term for numerator.
1dude,think this way.
tn = (3n - 1) / 3n = 1 - 1/3n .
now,do the summation u would get :n - [(3n - 1) / 3n.2]
1Q.1) its easy. just need to find out the sum and then compare.
HINT : take the odd terms i.e. 1 to (2n-1). add and subtract rest terms upto 2n, to this sum,and then use n-terms formula.