Aptitude - Numbers - Discussion
Discussion Forum : Numbers - General Questions (Q.No. 113)
113.
(22 + 42 + 62 + ... + 202) = ?
Answer: Option
Explanation:
(22 + 42 + 62 + ... + 202) = (1 x 2)2 + (2 x 2)2 + (2 x 3)2 + ... + (2 x 10)2
= (22 x 12) + (22 x 22) + (22 x 32) + ... + (22 x 102)
= 22 x [12 + 22 + 32 + ... + 102]
 |
Ref: (12 + 22 + 32 + ... + n2) = | 1 | n(n + 1)(2n + 1) |  |
|
| 6 |
| = |  |
4 x | 1 | x 10 x 11 x 21 |  |
| 6 |
= (4 x 5 x 77)
= 1540.
Discussion:
10 comments Page 1 of 1.
Harshita said:
3 years ago
Thank you for explaining the answer @Shweta.
Shweta said:
3 years ago
For given problem, use formula
-->Sum of squares of first n even numbers = 2n*(n+1)*(2n+1)/3.
If you want to find out
-->Sum of squares of n natural numbers = n*(n+1)*(2n+1)/6.
-->Sum of squares of first n odd numbers= n*(2n+1)*(2n-1)/3.
-->Sum of squares of first n even numbers = 2n*(n+1)*(2n+1)/3.
If you want to find out
-->Sum of squares of n natural numbers = n*(n+1)*(2n+1)/6.
-->Sum of squares of first n odd numbers= n*(2n+1)*(2n-1)/3.
(3)
JATIN SINGH said:
4 years ago
How 4 came? Explain me.
Saru said:
5 years ago
This formula is based on sum of squares of first positive integers=1^2+2^2+3^2+..........+10^2=n(n+1)(2n+1)/6.
Priya said:
7 years ago
Series is (2^2+4^2+6^2+.......).
= 2^2+(2*2)^2+(2*3)^2+.....+(2*10)^2
2^2(1+2^2+3^2+.......+10^2).
So it will be 4*(1/6)*n(n+1)(2n+1) where n= 10.
= 2^2+(2*2)^2+(2*3)^2+.....+(2*10)^2
2^2(1+2^2+3^2+.......+10^2).
So it will be 4*(1/6)*n(n+1)(2n+1) where n= 10.
Indra said:
7 years ago
I can't understand this. Can anyone explain this in detail?
Priya said:
7 years ago
Here the answer cannot be 1540. It should be 770 because 4 cannot be multiplied on it only 2 can be multiplied. Because 2^2 cannot be taken only 2 can be taken.
JECINTHA said:
10 years ago
Can anyone tell me what is the next number of the below series: -1,0,1,8,....?
Venu said:
10 years ago
I can't understand. Please explain?
Esha said:
1 decade ago
Did not understand.
Why 4 is multiplied with 1/6?
Why 4 is multiplied with 1/6?
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