Question

用数列的累加法做

已知数列an满足a(n+1）=an+3(n),a1=1.求an

Answer 1

a(n+1) = a(n) + 3^n,

累加法~~~
a(2) = a(1) + 3,
a(3) = a(2) + 3^2,
...
a(n-1) = a(n-2) + 3^(n-2),
a(n) = a(n-1) + 3^(n-1).

上面n-1个等式的等号两边分别累加，消去相同的和项，有，
a(n) = a(1) + 3 + 3^2 + ... + 3^(n-2) + 3^(n-1)
= 1 + 3 + 3^2 + ... + 3^(n-1)
= [3^n - 1]/(3-1)
= (3^n - 1)/2

Answer 2

a1=1,a2=a1+3*1,a3=a2+3*2……a(n+1)=an+3n
累加得a1+a2+a3+…+a(n+1)=1+a1+a2+…an+3*(1+2+…+n)=1+a1+a2+…+an+3n(n+1)/2
等式左边a(n+1)换成an+3n，左右两边同时减去a1+a2+…+an
得an+3n=1+3n(n+1)/2
an=3n(n+1)/2+1-3n