Aptitude - Numbers - Discussion
Discussion Forum : Numbers - General Questions (Q.No. 57)
57.
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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 1 of 2.
Richa said:
1 decade ago
Not getting it. Please explain more.
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.
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.
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)
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)
Vanitha said:
10 years ago
Can you please explain it clearly?
Madhur said:
9 years ago
More simple method.
Atul said:
9 years ago
I think it should be n+1/2.
Harsha vardhan said:
8 years ago
Given;
(1-1/n)+(1-2/n)+(1-3/n)+.....nterms.
We are seperating them as;
(1+1+1+1....n terms) - (1/n+2/n+3/n....n/n terms).
Now it becomes,
n-[1/n(1+2+3+....n terms) ] [taking 1/n as common].
Now it becomes;
n-[1/n(n(n+1)/2)] [sum of first N naturals numbers]
On simplification we get
n-(n+1)/2 [taking 2 as L.C.M] [ (2n-n-1)/2] ->->->->[(n-1)/2].
by taking 1/2 as common we get 1/2(n-1).
(1-1/n)+(1-2/n)+(1-3/n)+.....nterms.
We are seperating them as;
(1+1+1+1....n terms) - (1/n+2/n+3/n....n/n terms).
Now it becomes,
n-[1/n(1+2+3+....n terms) ] [taking 1/n as common].
Now it becomes;
n-[1/n(n(n+1)/2)] [sum of first N naturals numbers]
On simplification we get
n-(n+1)/2 [taking 2 as L.C.M] [ (2n-n-1)/2] ->->->->[(n-1)/2].
by taking 1/2 as common we get 1/2(n-1).
(10)
Veda said:
8 years ago
Given sum
= (1 + 1 + 1 + ... to n terms) - (1/n + 2 /n+ 3 /n+ ... to n terms)
There is another formula for sum of terms in A.P
Sum of terms n/2(a+1)
where a=first term that is 1/n
now substitute n/2(1/n+1)
we get( n+1)/2.
From the given question (1+1+1-------nterms) i.e, n.
We get n-(n+1/2).
After solving it we get n-1/2.
= (1 + 1 + 1 + ... to n terms) - (1/n + 2 /n+ 3 /n+ ... to n terms)
There is another formula for sum of terms in A.P
Sum of terms n/2(a+1)
where a=first term that is 1/n
now substitute n/2(1/n+1)
we get( n+1)/2.
From the given question (1+1+1-------nterms) i.e, n.
We get n-(n+1/2).
After solving it we get n-1/2.
(3)
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