How to solve this quadratic equation of one variable? 1.52 (x + 1) ^ 2 = 3.42 ^ 2 is square

How to solve this quadratic equation of one variable? 1.52 (x + 1) ^ 2 = 3.42 ^ 2 is square

1.52（x+1）^2=3.42
(x+1)^2=3.42/1.52=(2*3*3*19)/(2*2*2*19)=9/4
X + 1 = 3 / 2 or x + 1 = - 3 / 2
X = 1 / 2 or x = - 5 / 2

There is a three digit number, which is not only an even number, but also a multiple of nine, and can be divided by five. What is the minimum number of this three digit number

It is obvious from the title that the number is a multiple of the least common multiple of 2.9.5
SO 2 × 9 × 5 = 90,
So 90 × 1 = 90, 90 × 2 = 180
Exactly 180 is a three digit number

It is known that the sequence {CN} satisfies CN = 3 / bnxb (n + 1), BN = 3n-2. Find the first n terms and TN of the sequence {CN}

cn=3/[bnb(n+1)]=3/[(3n-2)(3(n+1)-2)]=3/[(3n-2)(3n+1)]=3×(1/3)×[1/(3n-2)-1/(3n+1)]=1/(3n-2)-1/[3(n+1)-2]Tn=c1+c2+...+cn=1/(3×1-2)-1/(3×2-2)+1/(3×2-2)-1/(3×3-2)+...+1/(3n-2)-1/[3(n+1)-2]=1/(3×1-2...

The sum of the first n terms of sequence BN is TN, 6tn = (3N + 1) BN + 2, find BN

When n ≥ 2, there is BN = tn-t (n-1), so from 6tn = (3N + 1) BN + 2 we get 6T (n-1) = (3 (n-1) + 1) B (n-1) + 2. By subtracting the two formulas above, we get 6 (tn-t (n-1) = (3N + 1) BN - (3n-2) B (n-1), namely 6bn = (3N + 1) BN - (3n-2) B (n-1), namely (3n-5) BN = (3n-2) B (n-1) BN / (3n-2) = B (n-1) / (3n-5) BN / (3n-2) = B

Find the first n terms of BN = 2 ^ (3n-2) and TN!

BN is an equal ratio sequence, B1 = 2, and the common ratio is 8
So we can directly replace the formula
2 (the nth power of 8 - 1)
tn=----------------------------
8-1

Sequence {an}, an = 9N (n + 1) / 2 ^ n for all positive integers n, if an ≤ m, then the range of M is_____________ .

An = 9N (n + 1) / 2 ^ Na (n + 1) = 9 (n + 1) (n + 2) / 2 ^ (n + 1)  a (n + 1) / an = (n + 2) / 2n = 1 / 2 + 1 / N  when n ≥ 2, a (n + 1) / an ≤ 1 / 2 + 1 / 2 = 1  an decreases monotonically when n ≥ 2. The maximum value of an = A2 = 9 × 2 × 3 / 2 &  178; = 27 / 2 > A1 = 9 × 2 / 2 = 9 sequence {an}, an = 9N (n + 1) / 2 ^ n for a

The sequence [an] satisfies (n + 1) an + 1 = 2 (n + 2) an + 3N2 + 9N + 6, A1 = 6, finding the general term formula

(n + 1) a (n + 1) = 2 (n + 2) an + 3N & # 178; + 9N + 6 (n + 1) a (n + 1) = 2 (n + 2) an + 3 (n + 1) (n + 2) both sides of the equation are divided by (n + 1) (n + 2) a (n + 1) / (n + 2) = 2An / (n + 1) + 3A (n + 1) / (n + 2) + 3 = 2An / (n + 1) + 6 = 2 [an / (n + 1) + 3] [a (n + 1) / (n + 2) + 3] / [an / (n + 1) + 3] = 2

The general term formula of sequence {an} is an = (n + 1) × 0.9n. Is there such a positive integer n
Let an ≤ an hold for any positive integer n? Prove the conclusion

It doesn't exist

Given that the sum of the first n terms of a sequence an is Sn = - N2 + 9N + 1, the general term formula of a sequence an is obtained

(1) When n = 1, A1 = S1 = - 1 + 9 + 1 = 9 (2) n ≥ 2, Sn = - N & # 178; + 9N + 1s (n-1) = - (n-1) &# 178; + 9 (n-1) + 1 = - N & # 178; + 11n-9  an = SN-S (n-1) = - 2n + 10N = 1, the general term formula which does not satisfy the above formula is segmented form an = {9N = 1 {- 2n + 10N ≥ 2

The first N-term product of sequence an is n-square, then the general term formula of an when n is greater than or equal to 2

∵a1*a2*a3*…… *an=n^2
When n ≥ 2, A1 * A2 * A3 * *a（n-1）=（n-1）^2
∴ an=[n/(n-1)]^2