1. The number one less than two times x is (); the sum of the square of a plus two times a is (); the sum of the sum of three a minus B is () (4) 1 / 3:1 / 20 = 1 and 7 / 9: X

1. The number one less than two times x is (); the sum of the square of a plus two times a is (); the sum of the sum of three a minus B is () (4) 1 / 3:1 / 20 = 1 and 7 / 9: X

（1）2x-1
(2) a^2+2a
(3)3a-b
(4) 20/3=(16/9)/x
x=16/9*3/20=4/15

What number is the square of 13

√13

C language how to calculate within 100 odd and odd add, even and even add?

#include "stdio.h"
void main()
{
int i=1,odd=0,even=0;
for(;i

Given the first n terms of arithmetic sequence {an} and Sn = - 2n2-n, find the expression of general term an

Sn=-2n2-N
Then sn-1 = - 2 (n-1) & sup2; - (n-1)
an=Sn-Sn-1
=-2n2-N - [-2(n-1)²-（n-1)]
=-2n2-n+2n2-4n+2+n-1
=-4n+1

All items of sequence {an} are positive numbers, and the sum of the first n items of Sn. For any n ∈ n *, there are always an, Sn and an2 in the arithmetic sequence. (1) find A1; & nbsp; (2) find the general term formula of sequence {an}; (3) let the sum of the first n items of sequence {BN} be TN, and BN = 1an2. For any positive integer n, there is always TN < 2

(1) It is known that for any n ∈ n *, there is always an, Sn and an2 in the arithmetic sequence, and each item of the sequence {an} is positive, and the solution is A1 = 1 (2). It is known that for n ∈ n *, there is always 2Sn = an + an2, which is proved that 〈 2sn-1 = an-1 + an-12 (n ≥ 2) and (2) 2An = an + an2-an-1-an-12 〉 an + an-1 = (an + an-1) (an-an-an-an-1) ∵ an, an-1 are positive, and ∵ an-an-an-an-1 = 1 (n ≥ 2) (3) BN = 1an2 = 1n2 ＜ 1n − 1-1n (n ≥ 2) when n = 1, TN = B1 = 1 ＜ 2; when n ≥ 2, TN = 112 + 122 + +1n2＜1+（1-12）+（12-13）+… +(1n − 1-1n) = 2-1n < 2. For any positive integer n, there is always TN < 2

It is proved that the sum of {an;} is {Sn;}
② If Sn = an & # 178; + BN + C (AC ≠ 0), please calculate whether {an} is still an arithmetic sequence

（1）
Sn=An^2+Bn
S(n+1)=A(n+1)^2+B(n+1)
S(n-1)=A(n-1)^2+B(n-1)
So a (n + 1) = s (n + 1) - Sn = an ^ 2 + 2An + A + BN + B-An ^ 2-BN = 2An + A + B
an=Sn-S(n-1)=An^2+Bn-An^2+2An-A-Bn+B=2An-A+B
So a (n + 1) - an = 2An + A + b-2an + A-B = 2A is a constant
Because a ≠ 0
So {an} is an arithmetic sequence with non-zero tolerance
（2）
If Sn = an & # 178; + BN + C
S(n+1)=A(n+1)^2+B(n+1)+C
S(n-1)=A(n-1)^2+B(n-1)+C
So a (n + 1) = s (n + 1) - Sn = an ^ 2 + 2An + A + BN + B + c-an ^ 2-bn-c = 2An + A + B
an=Sn-S(n-1)=An^2+Bn+C-An^2+2An-A-Bn+B-C=2An-A+B
So a (n + 1) - an = 2An + A + b-2an + A-B = 2A is a constant
{an} is an arithmetic sequence

It is known that the general term formula of {an} is an = 2N-1, and the general term formula of {BN} is BN = 3 ^ (n-1)

an*bn=(2n-1)×3^(n-1)Tn=a1b1+a2b2+a3b3+…… +anbnTn=1×3^0+3×3^1+5×3^2+…… +(2n-1)×3^(n-1) ①3Tn= 1×3^1+3×3^2+5×3^3+…… +(2n-3) × 3 ^ (n-1) + (2n-1) × 3 ^ n (2) from (1) - 2, - 2tn = 1 + 2 × [3 ^ 1 + 3 ^ 2 + +3^(n-1)]+(2...

If the general term formula of sequence {an} is an = NCOs (n π / 2), and the sum of the first n terms is Sn, then s2012 is equal to () a.1006 B.2012 c.503 d.0

My friend, this problem is a, the solution is as follows, n π 2 π∵ y = cos 2 period T = π = 4,2 ∵ can be divided into four groups of summation ＋a2009＝0,503－2－2010 a2 ＋a6 ＋… ＋a2010 ＝－2－6－… －2010＝ 2 ＝－503×1006,a3＋a7＋… ＋a2011＝0,5034＋2012 a4 ＋ a8 ＋… ＋ a2012 ＝ 4 ＋ 8 ＋… ＋ 2012 ＝ 2 ＝503×1008,∴S2012＝0－503×1006＋0×1008＝503－1006＋1008＝1006.

In the known sequence {an}, an = n2-n-50, then - 8 is its number ()
A. 5 items B. 6 items C. 7 items D. 8 items

Let an = n2-n-50 = - 8, we can get n2-n-42 = 0, the solution is n = 7 or n = - 6 (rounding), that is - 8 is the seventh term of the sequence

It is known that the sequence an is an increasing sequence, and its general formula is the square of an = n + λ n

A-an > 0
That letter is too hard to type. Change it to B
(n+1)^2+b(n+1)-n^2-bn>0
(n+1)^2-n^2+b>0
2n+1+b>0
b>-2n-1
Because n is a natural number that is not zero
And when n = 1, it can't form increasing sequence, because there is only one number
So when n = 2
-2 * 2-1 = - 5 Max
So b > - 5
That is, the value range of your letter is (- 5, positive infinity)