Aptitude - Numbers - Discussion
Discussion Forum : Numbers - General Questions (Q.No. 57)
57.
 |
1 - | 1 |  |
+ | |
1 - | 2 | |
+ | |
1 - | 3 | |
+ ... up to n terms = ? |
| n | n | n |
Answer: Option
Explanation:
| Given sum |
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Discussion:
15 comments Page 2 of 2.
Aratrika sengupta said:
1 decade ago
There is yet another method to solve the question
1+1+1....n times=n
n-1/n(1+2+3....)
n-1/n(n*(n+1)/2=n-1/2
you can ask if u have a doubt in my method.
1+1+1....n times=n
n-1/n(1+2+3....)
n-1/n(n*(n+1)/2=n-1/2
you can ask if u have a doubt in my method.
(2)
Lucky said:
1 decade ago
Hi,
We also solve in another method. This expresion (1/n + 2/n + 3/n + 4/n + ........ n terms)is write as 1/n(1+2+3+4+5+6+----+n)
Now apply the formula for sum of n natural numbers(n(n+1)/2).
then ,n-(1/n*n(n+1)/2)=n-(n+1)/2
=(2n-n-1)/2
=(n-1)/2
We also solve in another method. This expresion (1/n + 2/n + 3/n + 4/n + ........ n terms)is write as 1/n(1+2+3+4+5+6+----+n)
Now apply the formula for sum of n natural numbers(n(n+1)/2).
then ,n-(1/n*n(n+1)/2)=n-(n+1)/2
=(2n-n-1)/2
=(n-1)/2
(2)
Vamsi Krishna said:
1 decade ago
Sum of (1/n + 2/n + 3/n + 4/n + ........ n terms) is
Sum of all the terms which are in A.P is Sn = n/2 (2a+(n-1)d))
Using the above formula and substitute as n=n,a=1/n and d=1/n.
We will get as n/2(1+ 1/n).
If we subtract the above from 'n', we will get the result as (n-1)/2.
Sum of all the terms which are in A.P is Sn = n/2 (2a+(n-1)d))
Using the above formula and substitute as n=n,a=1/n and d=1/n.
We will get as n/2(1+ 1/n).
If we subtract the above from 'n', we will get the result as (n-1)/2.
Venumadhav said:
1 decade ago
Hai richa. In the given question substitute n=2 (for example). You get (1-1/2) =0. 5. Now cross check in the given options. 1/2(n-1) gives you the answer.
Richa said:
1 decade ago
Not getting it. Please explain more.
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