Question

设定义域为R的函数y=f(x)满足条件f(x+1)-f(x)=2x+1,且f(0)=0,数列{an}的前n项和Sn=f(

Answer 1

(1)

f(x+1)-f(x) =2x+1

f(n+1) -f(n) = 2n+1

f(n) -f(n-1) = 2n-1 (1)

.

...

f(2)-f(1) = 3 (n-1)

f(1)-f(0) = 1 (n)

(1)+(2)+...+(n)

f(n) -f(0) = 1+3+...+2n-1

=n^2

f(n) =n^2

Sn =f(n)

= n^2

(2)

f(x+1)-f(x) =2x+1

x=0

f(1)-f(0) =1

a1=S1= f(1) =1

an = Sn -S(n-1)

= 2n-1

Sn =f(n) = n^2