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.
Md. Saifuzzaman said:
8 years ago
Thank you @Majitha.
Beedisha said:
8 years ago
Why we are subtracting 1 to 10 square values?
Himansu mohanta said:
7 years ago
What is the value of n here, & how we know?
Venkat said:
7 years ago
I can't understand. Please help me to get it.
Tharakesh said:
7 years ago
Here, (1**2+2**2+3**2+.........+n**2) = (1/6)(n(n+1)(2n+1)
n=20.
(1/6)(20*21*41)=2870.
n=20.
(1/6)(20*21*41)=2870.
(1)
Nithish said:
7 years ago
How did ^66 Came? I didn't understand.
Gowry said:
7 years ago
It is very simple.
We can learn squares to addition.
121+144+169+196+225+256+289+324+361+400 = 2485.
But it is below 20^2.
We can learn squares to addition.
121+144+169+196+225+256+289+324+361+400 = 2485.
But it is below 20^2.
(2)
Sruthi said:
7 years ago
By applying n(n+1)(2n+1)/6 we get the answer directly.
Hence given upto 20.
So, here n=20.i.e.20(21)(41)/6=2870.
Hence given upto 20.
So, here n=20.i.e.20(21)(41)/6=2870.
(2)
VARSHA said:
6 years ago
Sn = S20 - S10.
S10 = (1/6)(n(n+1)(2n+1)
= (10*11*21)/6 =385.
S20 = (1/6)(n(n+1)(2n+1)
= (20*21*41)/6 = 2870.
Sn = S20-S10.
= 2870-385,
= 2485.
S10 = (1/6)(n(n+1)(2n+1)
= (10*11*21)/6 =385.
S20 = (1/6)(n(n+1)(2n+1)
= (20*21*41)/6 = 2870.
Sn = S20-S10.
= 2870-385,
= 2485.
(3)
Ruchita said:
6 years ago
Can anyone explain me formula?
I don't want to remember it but I want to understand it.
I want to understand formula like why we have to divide 6 in this formula?
I don't want to remember it but I want to understand it.
I want to understand formula like why we have to divide 6 in this formula?
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