High School Math Problems 22 26, High School Math Problems

s[n+1](s[n]+2)=s[n](2-s[n+1]) has s[n+1]s[n]=2(s[n+1]-s[n])=2b[n+1]s[n+1]s[n+1]s[n]=2b[n+1].

Find the reciprocal: b[n+1] s[n+1]s[n]=1 2 and s[n+1]-s[n]=b[n+1].

b[n+1] s[n+1]s[n]=1 s[n]-1 s[n+1]=1 2

This is the difference series.

The tolerance is -1 2

1/s[1]=1/b[1]=1

1/s[n]=3/2-n/2①

b[n]=s[n+1]-s[n]=1/(1-n/2)-1/(3/2-n/2)

1/2(1-n/2)(3/2-n/2)

2/(2-n)(3-n)

Little brother (little sister), I don't know how to ask teachers or classmates. This is a topic that is often encountered in the third year of high school.

By the question, open.

2sns(n+1)+2s(n+1)=2snso-[snsn(n+1)]=s(n+1)-snso,s(n+1)-sn

snsn(n+1)

This is ap d=1

1/s(n+1)- 1/sn = 1

1/sn- 1/s(n-1)=1

So, 1 s(n+1) -1 s1 = n So, s(n+1) = 1 (1+n).

sn=1/(n)

bn=-1/(n)(n-1)

a16=b6=-1/30

a18/a16=q2=4

q>0 so q=2

cn=bn*q^(n-1)

i.e. -[2 (n-1)].

-=cnn(n-1)

So, dn=-n*2 (n-1).

The standard split terms are summed, and the common ratio is 2

So. tn=(1-n)*(2^n) -1

I'm sorry if I make a mistake.

But the train of thought should be fine.

Because a2 and a5 are the two roots of the equation, Veda's theorem.

a2+a5=12

a2a5=27

That is, 2a1+5d=12 (1)(a1+d)(a1+4d)=27 (2) The solution gives d=2 (negative rounding) a1=1

An is an equal difference series, so an=1+(n-1)*2=2n-1tn=1-1 2bn (1)t(n-1)=1-1 2b(n-1) (2)(1)-(2) get bn=-1 2bn+1 2bn-1bn bn-1bn bn-1=1 3

bn is the first term is 2 3 common ratio is 1 3 proportional series bn = 2 3 * (1 3) n-1

So cn=(3 n * 2 3 * 1 3 n-1) ( 2n-1)(2n+1).

2 / (2n-1)(2n+1)

The sum of the items gives sn= 2 1*3 + 2 3*5 + 2 (2n-1)(2n+1).

1-1/3+1/3-1/5+1/5-1/7+ .1 / 2n-1 + 1 / 2n+1

1- 1 / 2n+1

2n / 2n+1

23 ｛x|x 1 or x 3}

25 minimum is 3 2

26 1/a+1/b=72

If you do it yourself, it's not necessarily right.

26-question process.

If the positive number a,b is full of 3+log2a=2+log3b=log6(a+b), let 3+log2a=2+log3b=log6(a+b)=x, then a=2 (x-3), b=3 (x-2), a+b=6 x1 a +1 b =a+b ab =72, and the other two questions will not be accepted.

If and only if the equation x+a=0 has no solution in (- 1), or if the solution x=-a--1 of x+a=0 is true.

So a (-1], should choose a