(simple operation, write the operation process) 100 + 99-98 + 97-96. + 3-2 + 1

(simple operation, write the operation process) 100 + 99-98 + 97-96. + 3-2 + 1

100+99-98+97-96…… +3-2+1
=100+（99-98）+（97-96）+（95-94）…… +（3-2）+1
=100+(1+1+1…… +1) There should be 49 ones in the middle of + 1
=100+1*49+1
=150
Equal ratio sequence a1 + A2 = 3, A2 + a3 = 6. A7 =?
A1 + A2 = 3, A2 + a3 = 6
a1+a1q=3① a1q+a1q2=6 ②a1q2-a1=3③
a1（1＋q）=3①a1（q2-1）=3③
The division of the two formulas leads to q = - 1 or 2. Because q is positive, q = 2 a = 1
a7=a1×q6=1×64=64
The N-1 power of 64,2
A math problem: love love love number = math, then love = () number = () learning = ()
Love △ number = mathematics, then love = (1) number = (3) learning = (7)
111÷3=37
A math problem: love love △ number = math, then love = (1) number = (3) learning = (7)
The math problems in grade three are so difficult
Love = (7) number = (14) learning = (14)
100+99-98-97+… +4+3-2-1=______ .
100+99-98-97+… +4+3-2-1，=（100+99-98-97）+… +(4 + 3-2-1), = 4 × 25, = 100
It is known that the increasing sequence {an} satisfies A2 + a3 + A4 = 28, and A3 + 2 is the median of the difference between A2 and A4. (I) find the general formula of the sequence {an}; (II) if BN = log2an + 1, find the first n terms and SN of the sequence {BN}
(I) let the common ratio of the equal ratio sequence {an} be Q. according to the meaning of the problem, there are 2 (A3 + 2) = A2 + A4, (1) and A2 + a3 + A4 = 28, substituting (1) into A3 = 8, so A2 + A4 = 20. Then there is a1q + a1q3 = 20a1q2 = 8, and the solution is A1 = 2q = 2 or A1 = 32q = 12, and {an} is increasing, so A1 = 2, q = 2. So an = 2n. (II) BN = log22n + 1 = n + 1. So Sn = N2 + 3N2
I love math
(I, AI, Shu and Xue represent one of the ten numbers 0-9 respectively)
I love + number + Learning = 10
I × love - number + Learning = 10
I love + number + Learning = 10
I × love × number × study =???
One hundred and twenty
I use x, y, a, B to express it. It's more convenient,
because
（1）x—y+a+b=10
（2）x×y—a+b=10
（3）x÷y+a+b=10
There are four unknowns in the question, but there are only three equations, which means that we can't find all the answers immediately, but it is also said that it is the number in 0-9, so there must be a small part of inference:
It can be seen from (1) (3): X-Y = x △ y (this is the key to solve the problem),
Suppose y is 1, X △ y = x, not equal to X-Y, so y is not equal to 1;
Suppose y is 2 and X can only be equal to 4;
Suppose y is 3, 3, 6 and 9;
If y is 4, 8 is not OK;
So y can only be 2!
So it's easy to solve, because y = 2, x = 4;
So:
（1）4—2+a+b=10
（2）4×2—a+b=10
（3）4÷2+a+b=10
So: a = 3, B = 5;
So: I × love × number × study = 4 × 2 × 3 × 5 = 120!
Done!
4-2+3+5=10
I × love × number × study =??? One hundred and twenty
I = 4, love = 2, number = 3, learning = 5,
I × love × number × study = 120
1 + 2 + 3 + 4 + 5 +... + 98 + 99 + 100 the sum of them is 2.10 + 20 + 30 +... + 80 + 90 + 100
1+2+3+4+5+...+98+99+100
=（1+100）*100/50
=5050
10+20+30+...+80+90+100
=(10+100)*10/2
=110*5
=550
It is known that a 2 + a 3 + a 4 = 28, a 3 + 2 is the mean of a 2 and a 4
Let the common ratio be Q. if the problem A2 + QA2 + Q ^ 2A2 = 28, A2 + Q ^ 2A2 = 2 (QA2 + 2) solves q = 2, A2 = 4, then an
= a1q ^ (n-1) = the nth power of 2
The common ratio of an is 2
The first is 2
Title: A3 = 8
∴a2+a4=20
And an = a1q ^ n-1
So 2q evaluates fan-5q + 2 = 0
Increasing the number sequence
So q = 2
So A1 = 2
So an = 2 ^ n
If A2 = 4, then A1 = 2, so a (n) = 2 ^ n seems to have a problem "{an} satisfies A2 + a3 + 2. It is 2 (A3 + 2) = A2 + A4 in the arithmetic mean of a2.a4, and 3a3 + 4 = 28 in the former formula,
2(a3+2)=a2+a4
a2+a3+a4=2(a3+2)+a3=3a3+4=28
a3=8;
a3=a2q=8 a2=8/q a4=a2q^2=8q
a2+a3+a4=28
8/q+8+8q=28
8/q+8q=20
Sorting 2q ^ 2-5q + 2 = 0
(2q-1) (Q-2) = 0, q = 1 / 2 (rounding off) or 2
So q = 2, A1 = 2; an = a1q ^ n-1 = 2 ^ n
A2 + a3 + A4 = 3A1 + 6D = 28, (A3 + 2) (A3 + 2) = a2xa4, (a1 + 2) (a1 + 2) = (a1 + D) (a1 + 3D), the first formula uses land to represent A1, and then substituting into the third formula can get D, and then D can calculate A1, so we can get the general formula an. I'm sorry
Multiply by 9 = learn numbers love me find out what numbers I love mathematics represent
I (1) love (0) count (8) learn (9) * 9 = learn (9) count (8) love (0) me (1)
1089*9=9801
1089*9=9801
Simple calculation (98 + 95 + 100 + 99 + 100 + 96 + 97 + 99)% 8
（98+95+100+99+100+96+97+99)%8
=(96+2 +96-1+96+4+96+3+96+4+96+96+1+96+3)%8
=(2-1+4+3+4+1+3)%8
=16%8
=0
96%8=0
So subtract 96 from each number
(2-1 + 4 + 3 + 4 + 0 + 1 + 3)% 8 = 0!