Aptitude - Numbers - Discussion

Discussion Forum : Numbers - General Questions (Q.No. 53)
53.
(112 + 122 + 132 + ... + 202) = ?
385
2485
2870
3255
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 1 of 4.

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.

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.

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.
(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.
(8)

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.

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).

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.

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.

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

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.
(3)


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