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# What is the result of the sum f(1)+...+f(20) if f(x)=x^2+2x+3

portoruj | Student, Grade 10 | eNoter

Posted January 23, 2011 at 11:26 PM via web

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What is the result of the sum f(1)+...+f(20) if f(x)=x^2+2x+3

Tagged with algebra1, math

giorgiana1976 | College Teacher | Valedictorian

Posted January 23, 2011 at 11:34 PM (Answer #1)

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We'll evaluate the sum by substituting x by the values 1,...,20.

f(1) = 1^2 + 2*1 + 3

f(2) = 2^2 + 2*2 + 3

f(3) = 3^2 + 2*3 + 3

......................

f(20) = 20^2 + 2*20 + 3

We'll calculate the sum:

f(1) + f(2) +..+ f(20) = (1^2+...+20^2) + 2*(1+...+20) + 3*20

We'll remove the brackets and we'll substitute the sum of the squares of the first 20 natural numbers, by the product:

1^2+...+20^2 = 20*(20+1)(2*20+1)/6

1^2+...+20^2 = 10*21*41/3

1^2+...+20^2 = 10*7*41

1^2+...+20^2 = 2870

We'll remove the brackets and we'll substitute the sum the first 20 natural numbers, by the product:

1+2+...+20 = 20(20+1)/2

1+2+...+20 = 10*21

1+2+...+20 = 210

f(1) + f(2) +..+ f(20) = 2870 + 2*210 + 3*20

f(1) + f(2) +..+ f(20) = 3350

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted January 23, 2011 at 11:37 PM (Answer #2)

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We know that f(x) = x^2 + 2x + 3

f(1) = 1^2 + 2*1 + 3

f(2) = 2^2 + 2*2 + 3

...

f(20) = 20^2 + 2*20 + 3

So the sum f(1) +f(2) ...f(20)

=> 1^2 + 2^2...+20^2 + 2*(1 + 2...+20) + 20 *3

=> 20*21*41 / 6 + 2*20*21/2 + 60

=> 2870+420+60

=> 3350

Therefore the required result is 3350.

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