# What is the formula to get the sum of the squares of the first ten positive numbers: 1^2 + 2^2 + ... + 10^2

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The formula for the sum of the first n integers is `S_n = (n*(n+1)*(2n+1))/6`

Adding the square of the first 10 integers gives: `S_10 = 1^2 + 2^2 +...+10^2` = `(10*(10+1)*(20 +1))/6` = 385

**The sum of the square of the first 10 integers is 385**

Formula is `S=(2n^3+3n^2+n)/6`

for `x=1 rArr S(1)=1`

for `x=2 rArr S(2)=5=1^2+2^2`

so that supose true fot n, we gonna show it true for every n>0 by induction:

Now `S(n+1)^2= S(n) + (n+1)^2`

that is:

`S(n+1)=(2n^3+3n^2+n)/6 +(n+1)^2`

`S(n+1)= (2n^3+3n^2+n)/6+n^2+2n+1`

`S(n+1)=(2n^3+3n^2+n+6n^2+12n+6)/6`

`S(n+1)=(2n^3+9n^2+13n+6)/6`

`S(n+1)=(2n^3+6n^2+6n+2+3n^2+7n+4)/6`

`S(n+1)=(2(n^3+3n^2+3n+1)+(3n^2+6n+3)+n+1)/6```

`S(n+1)=(2(n+1)^3+3(n^2+2n+1)+ (n+1))/6`

`S(n+1)=(2(n+1)^3+3(n+1)^2+(n+1))/6`

This prove formula is true for n+1 too

Then:

`S(10)=( 2(10)^3+3(10)^2+10)/6=` `(200+300+10)/6=2310/6=385```