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Algebra

Blazing goIITian

Joined:	13 Jul 2008
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25 Sep 2008 20:26:19 IST

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823

Another good problem(tricky)-Progressions

Engineering Entrance , JEE Main , JEE Main & Advanced , Mathematics , Algebra

Find the sum to n terms of the series .

Rates assured

Comments (9)

ravi teja

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25 Sep 2008 20:30:36 IST

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ans is n(n+1)/2+(6+n)/2+(1-n)/2n

rate me if my ans is correct.....

Rohit

Blazing goIITian

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25 Sep 2008 20:39:47 IST

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Sorry,but thats the wrong answer.

The correct ans is

Rohit

Blazing goIITian

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Posts: 382

25 Sep 2008 20:40:41 IST

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Helppppp pleaseee!!!!

ravi teja

Scorching goIITian

Joined: 22 Dec 2007

Posts: 293

25 Sep 2008 20:41:29 IST

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itz so easy...i think i have to simplify more....

ravi teja

Scorching goIITian

Joined: 22 Dec 2007

Posts: 293

25 Sep 2008 20:42:37 IST

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by generalising we can do....since if n=2 it is 4

if n=3 S=9..so we can take it as n2

i ll post soon the correct sol....

Rohit

Blazing goIITian

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25 Sep 2008 20:44:26 IST

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No ans is

Avirup Dasgupta

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25 Sep 2008 23:08:21 IST

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Rohit,

I am not sure if this is the correct method but you can try.

I will try to post the complete solution within 2 days.

Anant Kumar

Forum Expert

Joined: 10 Jul 2008

Posts: 598

25 Sep 2008 23:37:55 IST

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Let $r=1+\frac{1}{n}$

Then, the required sum

$S_n = 1 + 2r + 3r^2 + \ldots + n r^{n-1}$

Multiplying $S_n$ with $r$ and subtracting from $S_n$ , we obtain

$(1-r)S_n = 1+r+r^2 + \ldots + r^{n-1} - nr^n= \frac{1-r^n}{1-r} - n r^n$

Since $r=1+\frac{1}{n}$ , we get $1-r = -\frac{1}{n}$

Therefore, $-\frac{1}{n}\,S_n = -n (1-r^n)-nr^n = -n$

Hence, $\boxed{S_n = n^2}$

Rohit

Blazing goIITian

Joined: 13 Jul 2008

Posts: 382

26 Sep 2008 12:49:09 IST

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Avirup bhaiyya,
I couldn't understand your method

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