早教吧作业答案频道 -->其他-->
利用因式分解计算:(1)1002-992+982-972+…+42-32+22-12(2)1+24(52+1)(54+1)(58+1)•…•(532+1)(3)2n+4−2(2n)2(2n+2).
题目详情
利用因式分解计算:
(1)1002-992+982-972+…+42-32+22-12
(2)1+24(52+1)(54+1)(58+1)•…•(532+1)
(3)
.
22222222
24832
.
2n+4−2(2n) 2n+4−2(2n) 2n+4−2(2n)2n+4−2(2n)n+4−2(2n)2n)2n)n)2(2n+2) 2(2n+2) 2n+2)2n+2)n+2)
(1)1002-992+982-972+…+42-32+22-12
(2)1+24(52+1)(54+1)(58+1)•…•(532+1)
(3)
2n+4−2(2n) |
2(2n+2) |
22222222
24832
2n+4−2(2n) |
2(2n+2) |
2n+4−2(2n) |
2(2n+2) |
▼优质解答
答案和解析
(1)10022-9922+9822-9722+…+422-322+222-122
=(100+99)(100-99)+(98+97)(98-97)+…+(4+3)(4-3)+(2-1)(2+1)
=100+99+98+97+…+4+3+2+1
=101×50
=5050;
(2)1+24(522+1)(544+1)(588+1)•…•(53232+1)
=1+24×
×(52+1)(54+1)(58+1)•…•(532+1)
=1+564-1
=564;
(3)
=
=
.
52−1 52−1 52−12−152−1 52−1 52−12−1×(522+1)(544+1)(588+1)•…•(53232+1)
=1+56464-1
=56464;
(3)
=
=
.
2n+4−2(2n) 2n+4−2(2n) 2n+4−2(2n)n+4−2(2n)n)2(2n+2) 2(2n+2) 2(2n+2)n+2)
=
=
.
2n+1×8−2n+1 2n+1×8−2n+1 2n+1×8−2n+1n+1×8−2n+1n+12n+1×4 2n+1×4 2n+1×4n+1×4
=
.
7 7 74 4 4.
=(100+99)(100-99)+(98+97)(98-97)+…+(4+3)(4-3)+(2-1)(2+1)
=100+99+98+97+…+4+3+2+1
=101×50
=5050;
(2)1+24(522+1)(544+1)(588+1)•…•(53232+1)
=1+24×
52−1 |
52−1 |
=1+564-1
=564;
(3)
2n+4−2(2n) |
2(2n+2) |
=
2n+1×8−2n+1 |
2n+1×4 |
=
7 |
4 |
52−1 |
52−1 |
=1+56464-1
=56464;
(3)
2n+4−2(2n) |
2(2n+2) |
=
2n+1×8−2n+1 |
2n+1×4 |
=
7 |
4 |
2n+4−2(2n) |
2(2n+2) |
=
2n+1×8−2n+1 |
2n+1×4 |
=
7 |
4 |
2n+1×8−2n+1 |
2n+1×4 |
=
7 |
4 |
7 |
4 |
看了 利用因式分解计算:(1)10...的网友还看了以下:
已知Sn=2+22+23+…+2n(n∈N*),则Sn=. 2020-07-22 …
用数学归纳法证明等式:12-22+32+…+(2n-1)2-(2n)2=-n(2n+1)(n∈N* 2020-07-22 …
用数学归纳法证明等式:12-22+32-42+…+(2n-1)2-(2n)2=-n(2n+1)(n 2020-07-22 …
2+22+23+…+2n=(n为正整数).要过程! 2020-07-22 …
用数学归纳法证明“l+2+22+…+2n+2=2n+3-1,n∈N*”,在验证n=1时,左边计算所 2020-08-01 …
一元多项式在复数域内分解成一次因式的乘积(1)x^n-C(2n,2)x^(n-1)+C(2n,4) 2020-08-03 …
下列算式中正确的是()A.1+2+22+23+…+2n=11−2(1−2n)B.1−2+4−8+…+ 2020-10-31 …
1.(x+y)^4+x^4+y^4=2(x^2+xy+y^2)^22.(x-2y)x^3-(y-2x 2020-11-03 …
用数学归纳法证明:1+2+22+…2n-1=2n-1(n∈N)的过程中,第二步假设当n=k时等式成立 2020-12-05 …
设计求满足1+2+22+23+…+2n-1>10000的最小正整数n的程序框图,并编写相应的程序. 2021-01-15 …