(1) It is known that |ab-2| and |b- 1| are antonyms: |ab-2|=0 and |b- 1|=0 (only the inverse of 0 in non-negative numbers is 0), that is, b= 1 first.
1*2+2*3+3*4+4*5+……+2009*20 10
= 1*( 1+ 1)+2*(2+ 1)+3*(3+ 1)+4*(4+ 1)+…+2009*(2009+ 1)
=( 1^2+ 1)+(2^2+2)+(3^2+3)+(4^2+4)+…+(2009^2+2009)
=( 1+2+3+4+…+2009)+( 1^2+2^2+3^2+4^2+…+2009^2)
(2) The sum of1+2+3+4+…+2009 can be found by the basic formula, namely1+2+3+4+…+n = n (n+1)/2.
The key of this formula is to find the value of12+2 2+3 2+4 2+…+2009 2; The following is the derivation of the formula:
There are two methods here, let sn =12+2+...+n 2.
Method 1:
(it can be understood as 1 1+2 2+3 3+…+n N n).
1+2+3+4+5……+n
+2+3+4+5+……+n
3+4+5+……+n
4+5+……+n
……
+n
Use the summation formula:
( 1+n)n/2
+(2+n)(n- 1)/2
+……
+(n+n)(n-(n- 1))/2
Simplification = 0.5 * [(n+1) n+(n+2) (n-1)+(n+3) (n-2)+(n-4) (n-3)+... (n+n) ().
This is equivalent to getting an equation about Sn.
Simplify it:
n^3+n^2+ 1+(n+2)(n- 1)/2=3sn，
get
sn= 1/3*n^3+ 1/2*n+ 1/6*n= 1/6*n(n+ 1)(2n+ 1)
Method 2:
Sn=S(n- 1)+n^2
=s(n- 1)+ 1/3*[n^3-(n- 1)^3]+n- 1/3
=s(n- 1)+ 1/3*[n^3-(n- 1)^3]+ 1/2*[n^2-(n- 1)^2]+ 1/6
=s(n- 1)+ 1/3*[n^3-(n- 1)^3]+ 1/2*[n^2-(n- 1)^2]+ 1/6*[n-(n- 1)]
That is, sn-1/3 * n3-1/2 * N2-n/6 = s (n-1)-1/3 * (n-1) 3-65448.
All right! The left side of the equation is all n, and the right side is all (n- 1). If we continue recursion, we can get
sn- 1/3*n^3- 1/2*n^2-n/6
=s(n- 1)- 1/3*(n- 1)^3- 1/2*(n- 1)^2-(n- 1)/6
=s(n-2)- 1/3*(n-2)^3- 1/2*(n-2)^2-(n-2)/6
……
=s( 1)- 1/3*( 1- 1)^3- 1/2*( 1- 1)^2-( 1- 1)/6
=0
So sn =1/3 * n3+1/2 * n+1/6 * n.
So the original formula = 2009 * (2009+1)/2+2009 * (2009+1) * (2 * 2009+1)/6 = 270686330.
The derivation of the second formula of this problem is quite difficult for students.
Finally, I will help the landlord! !