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 1 of 2.
Shreya said:
3 years ago
I am not understanding it. Please, anyone, help me to get this.
(4)
Kelbin said:
4 years ago
=1+2+3+.........+n=n(n+1)/2.
= n-1/n(1+2+3+....+n)
= n-1/n *n(n+1)/2,
= n-(n+1)/2,
= 2n-n-1/2,
= n-1/2.
= 1/2(n-1).
= n-1/n(1+2+3+....+n)
= n-1/n *n(n+1)/2,
= n-(n+1)/2,
= 2n-n-1/2,
= n-1/2.
= 1/2(n-1).
(8)
Sujata said:
4 years ago
Please anyone explain me, I can't understand.
(3)
MADHU. said:
5 years ago
Please kindly tell me how to find 9it? Anyone explain to me, please.
Gowry said:
7 years ago
Can anyone explain clearly? Please.
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)
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)
Atul said:
9 years ago
I think it should be n+1/2.
Madhur said:
9 years ago
More simple method.
Vanitha said:
10 years ago
Can you please explain it clearly?
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