早教吧作业答案频道 -->其他-->
设f(i,k)=i•2(k-1)(i∈N*,k∈N*),如f(2,3)=2×2(3-1)=8.对于正整数m,n,当m≥2,n≥2时,设g(i,n)=f(20,n)+f(21,n)+…+f(2i,n),S(m,n)=mi=1(-1)ig(i,n),则S
题目详情
设f(i,k)=i•2(k-1)(i∈N*,k∈N*),如f(2,3)=2×2(3-1)=8.对于正整数m,n,当m≥2,n≥2时,设g(i,n)=f(20,n)+f(21,n)+…+f(2i,n ),S(m,n)=
(-1)ig(i,n),则S(4,6)=______.
m |
i=1 |
▼优质解答
答案和解析
由f(i,k)=i•2(k-1),
得g(i,n)=f(20,n)+f(21,n)+…+f(2i,n )
=20×2n-1+21×2n-1+22×2n-1+…+2i×2n-1
=(1+2+22+…+2i)•2n-1
=
•2n−1
=(2i+1-1)•2n-1.
又S(m,n)=
(-1)ig(i,n),
∴S(4,6)=(-1)1•(22-1)•25+(-1)2•(23-1)•25+(-1)3•(24-1)•25+(-1)4•(25-1)•25
=(-3+7-15+31)×32=640.
故答案为:640.
得g(i,n)=f(20,n)+f(21,n)+…+f(2i,n )
=20×2n-1+21×2n-1+22×2n-1+…+2i×2n-1
=(1+2+22+…+2i)•2n-1
=
1×(1−2i+1) |
1−2 |
=(2i+1-1)•2n-1.
又S(m,n)=
m |
i=1 |
∴S(4,6)=(-1)1•(22-1)•25+(-1)2•(23-1)•25+(-1)3•(24-1)•25+(-1)4•(25-1)•25
=(-3+7-15+31)×32=640.
故答案为:640.
看了设f(i,k)=i•2(k-1...的网友还看了以下:
定义数列{an}:a1=1,当n≥2时,an=an−1+r,n=2k,k∈N*2an−1,n=2k 2020-05-13 …
MATLAB 如何赋值 画实部虚部图像 画图clearclcsyms a j k n u e A 2020-05-16 …
数列{an}满足a1=1,an+1=2^n+1*an/an+2^n(n∈N+)1)证明:数列{2^ 2020-05-17 …
已知各项全不为零的数列{ak}的前k项和为Sk,且Sk=akak+1(k∈N*),其中a1=1。( 2020-07-20 …
已知数列{an}满足a1=1,a2=3,且an+2=(1+2|cosnπ2|)an+|sinnπ2| 2020-10-31 …
A{n│n=2k+1,k∈Z}、B{m│m=2l-1,l∈Z}如果n∈A,那么存在k∈Z,使n=2k 2020-10-31 …
an=n,当n≥2,{bn}满足b(k+1)/bk=(k-n)/a(n+1)(k=1,2…n-1), 2020-10-31 …
已知数列满足:a1=1,an+1=,(n∈N*),若bn+1=(n-λ),b1=-λ,且数已知数列满 2020-11-19 …
(2009•上海模拟)已知数列{an}满足a1=25,且对任意n∈N*,都有2an-2an+1=3a 2020-12-09 …
已知数列{an}满足a1=25,且对任意n∈N*,都有anan+1=4an+2an+1+2.(Ⅰ)求 2020-12-09 …