Can you find the remainder?

This is an alternate solution.

Here, we need to find

$1^{2} + 2^{2} + 3^{2} + ..... + 2013^{2}$ mod $7$

Now according to the formula of squares of first $n$ natural numbers,

$1^{2} + 2^{2} + 3^{2} + ..... + n^{2} = \frac{n \times (n + 1) \times (2n + 1)}{6}$

Therefore, we need to find

$= \frac{2013 \times (2013 + 1) \times (2 \times 2013 + 1)}{6}$ mod $7$

$= \frac{2013 \times 2014 \times 4027}{6}$ mod $7$

$= \frac{671 \times 2014 \times 4027}{2}$ mod $7$

$= 671 \times 1007 \times 4027$ mod $7$

$=(671$ mod $7 \times 1007$ mod $7 \times 4027$ mod $7$ ) mod $7$

$= -1 \times -1 \times 2$ mod $7$

$= 2$ mod $7$

$= \boxed{2}$

That's the answer!

let see the pattern :

$1^{2}$ mod 7 = 1

$2^{2}$ mod 7 = 4

$3^{2}$ mod 7 = 2

$4^{2}$ mod 7 = 2

$5^{2}$ mod 7 = 4

$6^{2}$ mod 7 = 1

$7^{2}$ mod 7 = 0

this pattern will be repeated for each seven numbers in a sequence.But, for the last sequence there are just 4 numbers.,

so, the sum for a sequence = 1 + 4 + 2 + 2 + 4 + 1 + 0 = 14 there are 287 sequences that have 7 numbers, so the sum is = 14 * 287 = 4018 the sum of the last four numbers is = 1 + 4 + 2 + 2 = 9.

so, the total is = 4018 + 9 = 4027, with this total we can calculate the remaining = 4027 mod 7 = 2

Observe that the sum above can be written as $\frac{(2013)(2014)(4027)}{6} = \frac{2013}{3} \times\frac{2014}{2} \times 4027$ . So, we have

$\frac{2013}{3} = 671 = 11\times61 \overline= 4\times5 \overline= 6\pmod7$

$\frac{2014}{2} = 1007 = 19\times53 \overline= 5\times4 \overline= 6\pmod7$

$4027 \overline= 2\pmod7$ (We don't prime factorize it since it is prime)

Therefore,

$\frac{(2013)(2014)(4027)}{6} \overline= 6\times6\times2 \overline= 2\pmod7$

Epilogue: Alright, this is a simple proof why $\Sigma^n_{k=1} k^2 = \frac{n(n+1)(2n+1)}{6}$

Let $u_k = k^3$ , we have

$u_{k+1} - u_k = (k+1)^3 - k^3$

$u_{k+1}-u_k = 3k^2+3k+1$

Then,

$\Sigma^n_{k=1} (u_{k+1} - u_k) = \Sigma^n_{k=1} (3k^2+3k+1)$

Telescoping LHS,

$u_{n+1} - u_1 = 3\Sigma^n_{k=1} k^2 + 3\Sigma^n_{k=1} k + \Sigma^n_{k=1} 1$

Plug in the identity $\Sigma^n_{k=1}k = \frac{n(n+1)}{2}$ ,

$(n+1)^3 -1^3 = 3\Sigma^n_{k=1} k^2 + 3\frac{n(n+1)}{2} + n$

$n^3 +3n^2 +2n =3\Sigma^n_{k=1} k^2 + 3\frac{n(n+1)}{2}$

Multiplying 2 on both sides,

$2n^3 +6n^2 + 4n = 6\Sigma^n_{k=1} k^2 + 3n^2 +3n$

$2n^3+ 3n^2 +n =6\Sigma^n_{k=1} k^2$

$\Sigma^n_{k=1} k^2 = \frac{2n^3+ 3n^2 +n }{6}$

Factorize the RHS, we have

$\boxed{\Sigma^n_{k=1} k^2 =\frac{n(n+1)(2n+1)}{6}}$

We know if 1 + 2 + 3 + ... + n = 1/2 . n ( n + 1). Then, we substituting ; <--> (n+1)^3 = 1 + 3(1^2 + 2^2 + 3^2 + ... + n^2) + 3 n (n + 1)/2 + nx1 ---> n^3 + 3.n^2 + 3n + 1 = 1 + 3(1^2 + 2^2 + 3^2 + ... + n^2) + 3/2 .n^2 + 3n/2 + n ---> n^3 + 3/2 . n^2 + n/2 = 3 (1^2 + 2^2 + 3^2 + ... + n^2) ---> (1^2 + 2^2 + 3^2 + ... + n^2) = n/6 (2n^2 + 3n + 1) = 1/6 . n(n+1)(2n+1). So, now we've that 1^2 + 2^2 + 3^2 + ... + n^2 = 1/6 . n(n+1)(2n+1). So, the problem above is (1/6 . 2013 . 2014 . 4027 ) mod (7) = 2. Answer : 2

2^2 - 1^2 = 3

3^2 - 2^2 = 5

4^2 - 3^2 = 7

n=2

since the given is from 1^2 up to 2013^2

then 2013 x 2 = 2046

2046 + 1 = 2047

2047 / 7 = 575 with a remainder of 2

(n(n+1) (2n+1))/6 = sum ((2013)(2014)(2 2013+1))/6 = 2721031819 , 2721031819/7 reminder =2

yo thats simple

$1^2$ + $2^2$ + $3^2$ +....+ $n^2$ = $\frac{n}{6}$ (n+1)(2n+1)

Therefore

$1^2$ + $2^2$ + $3^2$ +....+ $2013^2$

= $\frac{2013}{6}$ (2013+1)(4026+1)

= (671)(1007)(4027)

= 2721031819

Dividing it by 7, we get the remainder as 2.

If anyone has a shorter and easier solution, kindly tell me

Using the formula for the sum of squares series till n terms=n(n+1)(2n+1)/6 , = (2013 X 2014 X 4027) / (6 X 7) =(2013 X 2014 X 4027) / (42),

Taking individual remainders of the numerator by 42,

= (39 X 40 X 37 ) /42 =57720/42 =1374 =2(MOD 42)

ANSWER IS 2.

$\sum_{k=1}^nk^2=\frac{n(n+1)(2n+1)}{6}$ . Therefore $1^2+2^2+\cdots+2013^2=\frac{2013\times2014\times4027}{6}=671\times1007\times4027$ . Now $671\equiv-1~(\mod~7),~1007\equiv-1~(\mod~7),~4027\equiv 2~(\mod~7)$ .

sum of squares of n natural numbers is given by

= [n(n+1)(2n+1)]/6

Here n = 2013

= 2013 x2014 x 4027/6

= 16326190914/6

=2721031819/7

=Quotient 388718831 and remainder =2

Using the formula of sum of squares we get 1^2 + 2^2 ... + 2013^2=2013x2014x4027/6 which is cong to 2(mod 7)