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 3 of 4.
Abhinav Trivedi said:
10 years ago
We know that Sum of 1 square to 20 square = 2870 by the formula n/6 X (n+1) X (2n+1).
Therefore sum of 1 square to 10 square = 385.
So, the sum of 11 square to 20 square is 2870 - 385 = 2485.
Therefore sum of 1 square to 10 square = 385.
So, the sum of 11 square to 20 square is 2870 - 385 = 2485.
Inder said:
10 years ago
From where this come from (12 + 22 + 32 +.....+ 202) - (12 + 22 + 32 +.....+ 102).
Monika said:
1 decade ago
This sum is very easy with this formula n (n+1)*(2n+1) divide by 6 20(20+1)*(20*2+1) divided by 6 after multiplying break squares and divided with 6 then no. after multiplying dividing with 6 and subtract of both answer.
Ganesh NB said:
1 decade ago
@Kavitha.
According to the formula we find for the whole.
1 square to 20 square = 2870.
But we need only 11 square to 20 square.
So we subtract 1 square to 10 square = 385.
Hence 11 square to 20 square = 2870-385 = 2485.
According to the formula we find for the whole.
1 square to 20 square = 2870.
But we need only 11 square to 20 square.
So we subtract 1 square to 10 square = 385.
Hence 11 square to 20 square = 2870-385 = 2485.
Kavitha chowdary said:
1 decade ago
According to the formula the answer is 2870. But why we subtract 385? Anyone please tell me.
Reneesh said:
1 decade ago
The method is simple,
Firstly we need to find out the squares from 1 to 20 instead of 11 to 20.
Then we can easily find out the squares of 11^2.20^2 by taking the difference between the (1^2+2^2+.20^2) - (1^2+2^2+.10^2).
Firstly we need to find out the squares from 1 to 20 instead of 11 to 20.
Then we can easily find out the squares of 11^2.20^2 by taking the difference between the (1^2+2^2+.20^2) - (1^2+2^2+.10^2).
Abhimanyu jakhar said:
1 decade ago
Simple method is:
n/2[firstnum + lastnum]-120.
n = 50{totalnum}.
n/2[firstnum + lastnum]-120.
n = 50{totalnum}.
Bijoy said:
1 decade ago
Where did we get the value of {n(n+1)(2n+1)}/6 from?
Can someone help please in simple way?
Can someone help please in simple way?
Sumathi said:
1 decade ago
How do we get the value of {n(n+1)(2n+1)}/6?
Can please anyone tell me?
Can please anyone tell me?
Varun said:
1 decade ago
Its simple sum of squares of natural number till 20 minus sum till 10.
Formula to calculate such sum is
{n(n + 1)(2n + 1)}/6. n is the lats digit
ie. 20 and 10 in this example
Formula to calculate such sum is
{n(n + 1)(2n + 1)}/6. n is the lats digit
ie. 20 and 10 in this example
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