Aptitude - Numbers - Discussion
Discussion Forum : Numbers - General Questions (Q.No. 53)
53.
(112 + 122 + 132 + ... + 202) = ?
Answer: Option
Explanation:
(112 + 122 + 132 + ... + 202) = (12 + 22 + 32 + ... + 202) - (12 + 22 + 32 + ... + 102)
 |
Ref: (12 + 22 + 32 + ... + n2) = | 1 | n(n + 1)(2n + 1) |  |
|
| 6 |
| = |  |
20 x 21 x 41 | - | 10 x 11 x 21 |  |
| 6 | 6 |
= (2870 - 385)
= 2485.
Discussion:
33 comments Page 4 of 4.
Gauri Sharma said:
5 years ago
I didn't got it. Please explain.
(2)
Sagar jeevtani said:
5 years ago
@All.
To find the sum of (square of natural numbers) we can use the formula that is.
sn=n (n+1) (2n+1) /6 upto n terms.
But here,We have to find the sum of the square of a natural number between 11 to 20 then we have to subtract the sum of the square of natural number from 1 to 10 then we can get the answer.
To find the sum of (square of natural numbers) we can use the formula that is.
sn=n (n+1) (2n+1) /6 upto n terms.
But here,We have to find the sum of the square of a natural number between 11 to 20 then we have to subtract the sum of the square of natural number from 1 to 10 then we can get the answer.
(3)
J shivani said:
5 years ago
Here, we use n[n+1][2n+1]/6.
Here we take n=20 then substitute we get 2870
Next we have to take n=10 then substitute we get 385
Later Subtract 2870-385=2485.
It is bcoz in n=20 we are considering squares from 1 to 20 but it is from 11 to 20.
So, t n=20 then the sum of 1 to 20 is subtracted.
Here we take n=20 then substitute we get 2870
Next we have to take n=10 then substitute we get 385
Later Subtract 2870-385=2485.
It is bcoz in n=20 we are considering squares from 1 to 20 but it is from 11 to 20.
So, t n=20 then the sum of 1 to 20 is subtracted.
(8)
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