Sorry, an error occurred while loading the content.

## Re: Number System Problems

Expand Messages
• Here goes the soln 1)It can be seen that any term in the series can be written as Tn=(2n+1)/(n(n+1))^2 2n+1=(n+1)^2-n^2 Tn=n^-2 - (n+1)^-2 Summation to n terms
Message 1 of 3 , Jul 30, 2003
Here goes the soln
1)It can be seen that any term in the series can be written as
Tn=(2n+1)/(n(n+1))^2
2n+1=(n+1)^2-n^2
Tn=n^-2 - (n+1)^-2
Summation to n terms is
Sn= 1-(n+1)^-2
putting n=9 to the given problem
We get S9=99/100

2)Tn for this series is
Tn=1/n - 1/(n+1)
Sn=1-1/(n+1)
X=Sn=n/(n+1)

3) It can be seen that

7^1=7
7^2=49
7^3=...3
7^4=...1
7^5=...7
7^5 ends in 7.so also 7^5*7^5=7^25.
In general 7^5^n ends in 7.for given prob n=6^13.

--- In ascent4cat@yahoogroups.com, "Tarsem Saini" <ts_chd@y...> wrote:
> Hi All,
> Please help me out to solve the following problems in CAT approach.
>
> 1)find the value of 3/4 + 5/36 + 7/144 + ..... + 17/5184 + 19 / 8100
>
> a) 0.96
> b) 0.95
> c) 0.99
> d) None of These
>
> /******************************************************************/
>
> 2)Compute x if x = 1/1*2 + 1/2*3 + 1/3*4 + ....+ 1/(n-1)*n + 1/n*
(n+1)
>
> a) n/(n-1)
> b) 2n/(n+1)
> c) n/(n+1)
> d) None of these
>
> /******************************************************************/
>
> 3) What is the Unit digit of 7^5^6^13 ?
>
> I have been able to solve this question by calculating power and
then
> devide by 4 .But it take some time to calculate power. Please tell
> the Shortcut for this..
>
> /******************************************************************/
>
> Take care and prepare well.
>
> Saini
• hi 1 . the general form of the nth term is 2n + 1 = 2n + 1 = 1 [ 1 - n^2 ] = 1 - 1 (n(n+1))^2 n^2(n^2+2n+1) n^2
Message 2 of 3 , Jul 30, 2003
hi
1 . the general form of the nth term is
2n + 1    2n + 1            = 1     [ 1 - n^2       ] = 1      -      1
(n(n+1))^2   n^2(n^2+2n+1)     n^2        (n+1)^2       n^2       (n+1)^2
now as the value of n ranges from 1 to 9
the given sum is  1 - 1    +  1     -  1     ......... 1      -      1
2^2     2^2     3^2            9^2       10^2
=1 -   1     =0.99
100
so Ans - c)
2. 1          =  1     -       1
n(n+1)       n          (n+1)
here also as n ranges from 1 to n
1 -  1 + 1  - .....  -    1    =  1 -    1      =   n
2    2              n+1            n+1        n+1
so Ans - c)

3. anyway 5^6^13 or even 5^25000 has same last 2 digits ie 25
so the qn becomes smething like
7^((n * 100) + 25) so this becomes 7^(4m +1)
this gives a 7 in the units place

bye
g'luk everyone

Tarsem Saini <ts_chd@...> wrote:
Hi All,
Please help me out to solve the following problems in CAT approach.

1)find the value of 3/4 + 5/36 + 7/144 + ..... + 17/5184 + 19 / 8100

a) 0.96
b) 0.95
c) 0.99
d) None of These

/******************************************************************/

2)Compute x if x = 1/1*2 + 1/2*3 + 1/3*4 + ....+ 1/(n-1)*n + 1/n*(n+1)

a) n/(n-1)
b) 2n/(n+1)
c) n/(n+1)
d) None of these

/******************************************************************/

3) What is the Unit digit of 7^5^6^13 ?

I have been able to solve this question by calculating power and then
devide by 4 .But it take some time to calculate power. Please tell
the Shortcut for this..

/******************************************************************/

Take care and prepare well.

Saini

------------------------ Yahoo! Groups Sponsor ---------------------~-->
Free shipping on all inkjet cartridge & refill kit orders to US & Canada. Low prices up to 80% off. We have your brand: HP, Epson, Lexmark & more.
http://www.c1tracking.com/l.asp?cid=5510
http://us.click.yahoo.com/GHXcIA/n.WGAA/ySSFAA/DxDolB/TM
---------------------------------------------------------------------~->

Ascent Education
An IIM Alumni Venture
Class for CAT, XAT, GRE, GMAT
http://www.ascenteducation.com

Archives of past CAT questions can be viewed at http://www.ascenteducation.com/india-mba/iim/cat/questionbank/questionbank.shtml

To unsubscribe from this group, send an email to:
ascent4cat-unsubscribe@yahoogroups.com

Your use of Yahoo! Groups is subject to http://docs.yahoo.com/info/terms/