What is the single digit when the square of 1 is added to the square of 2011?

You look for patterns, 1 2 + 2 2 +...9^2+10^2+..2010^2+2011^2

1 2 + 2 2 + 3 2 + 4 2 = 1 + 4 + 9 + 16 single digits are 0, which is a multiple of 10.

6 2+7 2+8 2+9 The 2 single digits are also 0, which is a multiple of 10.

Only 5 2 single digits are squared ,,, multiples of 510 Needless to say, the single digit is 0 Have you found a pattern? Between 1-10 when they are squared, the single digit is 5

Before 2010, there were 201 5's in the single digit, and there would be a 5 in the single digit, and in 2011, 2 digits were 1

So the answer is 6.

Discovery: The single digits are: (1,4,9,6,5,6,9,4,1,0),(1,4,9,6,5,6,9,4,1,0)......cycle, the cycle is 10

Because there are only single digits, you only need to add them up.

1 to 2011 2 has 2011 numbers, i.e. 201 periods + 1, i.e., the single digit is 6

If you don't understand, please ask.

The size of the single digit is only related to the square digit of the number, and 1 2 + ......The 2010 2-digit change is just 1 2+ ......10 2 of the single digit 5 of 201 times is added, so 1 2 + ......The single digit of 2011 2 is 5+1=6.

1 2 + 2 2 + 3 2 + ......n^2=n*(n+1)(2n+1)/6

1^2+2^2+3^2+……Filial piety +2006 2=2006(2006+1)(2*2006+1) 6=2692751091

So the single digit is 1

Added: A group of ten, because of the annihilation of the Feng teasing for ten clans to sell a cycle. 2000 10 = 200, there are 6 bases left.

1+4+9+6+5+6+9+4+1+0+1+2+9.。。File annihilation...

For the stupid only Dapei 1

Memorize the formula.

1^2+2^2+3^2+……2n-1)^2= 1/6 *n(n+1)(2n+1)

In this case, 2n-1=2015, n=1008

Substituting the formula gets.

Use the formula 1 2 + 2 2 + 3 2....+n^2

1 6 *n(n+1)(2n+1) then gets.

1 6 * 2017 * 2018 * (2017 * 2 +1) Obviously 2017 * 2 + 1 = 4035

If there is 5, then the single digit must be 5

Apply the sum of squares formula.

1^2+2^2+3^2+…+n^2=n(n+1)(2n+1)/6n(n+1)(2n+1)/6

Just put n=2007 in it.