Aptitude - Numbers - Discussion

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.

@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.

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.

I didn't got it. Please explain.

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.

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.

Here, (1**2+2**2+3**2+.........+n**2) = (1/6)(n(n+1)(2n+1)
n=20.
(1/6)(20*21*41)=2870.

I can't understand please explain me clearly.

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.

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.