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 2 of 4.
Himansu mohanta said:
7 years ago
What is the value of n here, & how we know?
Beedisha said:
8 years ago
Why we are subtracting 1 to 10 square values?
Md. Saifuzzaman said:
8 years ago
Thank you @Majitha.
Atharva said:
8 years ago
Thanks, I got it.
Majitha said:
8 years ago
Numbers which starts from 1 then the sum of series formula = 1/6* n(n + 1)(2n + 1).
But in our sum which starts from 12.
So we have to find two separate sum of series.
1st one for 1 to 10.
2nd one for 11 to 20.
As per our formula,
For 1 to 10 here.
N = 10
So we get,
1/6 * 10(10+1)(2*10+1) = 385.
Like as n=20.
1/6* (20(20+1)(2*20+1) = 2485.
We have to find ans for (11^2 + 12^2 + 13^2 + ... + 20^2).
So do sub of 2485 -385 = 2485.
But in our sum which starts from 12.
So we have to find two separate sum of series.
1st one for 1 to 10.
2nd one for 11 to 20.
As per our formula,
For 1 to 10 here.
N = 10
So we get,
1/6 * 10(10+1)(2*10+1) = 385.
Like as n=20.
1/6* (20(20+1)(2*20+1) = 2485.
We have to find ans for (11^2 + 12^2 + 13^2 + ... + 20^2).
So do sub of 2485 -385 = 2485.
Anu said:
8 years ago
I can't understand please explain me clearly.
StK said:
9 years ago
It's very easy.
All u have to do is learn the formula.
(1^2 + 2^2 + 3^2+.......n^2) = 1/6 n(n+1)(2n+1).
So, we know the solution of above type of series so we can reduce the series given in Qns to this type of series.
So, what we are doing next is expanding the given series of question into (1^2 + 2^2+....+20^2), which will be solved by above formula 1/6 n(n+1)(2n+1) with the value n=20 and we will get 2870.
But the series in the question is only of squares of 11 to 20 so, now we will find the solution by using the same formula for squares of 1 to 10 ( by taking n=10 for this series as we are solving only up to 10 ^2) and we will get a solution of 385.
After that, we will minus 385 from 2870 and we will find a solution for the given series as 2485.
All u have to do is learn the formula.
(1^2 + 2^2 + 3^2+.......n^2) = 1/6 n(n+1)(2n+1).
So, we know the solution of above type of series so we can reduce the series given in Qns to this type of series.
So, what we are doing next is expanding the given series of question into (1^2 + 2^2+....+20^2), which will be solved by above formula 1/6 n(n+1)(2n+1) with the value n=20 and we will get 2870.
But the series in the question is only of squares of 11 to 20 so, now we will find the solution by using the same formula for squares of 1 to 10 ( by taking n=10 for this series as we are solving only up to 10 ^2) and we will get a solution of 385.
After that, we will minus 385 from 2870 and we will find a solution for the given series as 2485.
Sumedha said:
9 years ago
I didn't get a proper answer. Please, can anybody give me proper answer & shortcut.
Somu said:
9 years ago
Nice explanation @Kavitha.
Tareque said:
9 years ago
10 * 11 * 21 / 6 How it came?
Please explain.
Please explain.
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