Given that the image of the function y = MX - (4m-3) (M is a constant) passes through the origin, what is the value of M

Given that the image of the function y = MX - (4m-3) (M is a constant) passes through the origin, what is the value of M

Passing through the origin, that is, x = 0, y = 0
So 0 = 0 - (4m-3)
m=3/4

Given that the sum of the first n terms of the sequence {an} is Sn, Sn = 2n ^ 2-3n + 6, then the general term formula of the function is an=

an = Sn-S(n-1)
= 2n^2-3n+6-[2(n-1)^2-3(n-1)+6]
= 2n^2-3n+6-[2n^2-4n+2-3n-3+6]
= 4n+1
When n = 1
s1=a1=5
Therefore, the general formula of sequence is as follows:
4n+1

Let A1 = 2, point (an, an + 1) be on the image of function f (x) = x2 + 2x, where n = 1, 2, 3 (I) prove that the sequence {LG (1 + an)} is an equal ratio sequence; (II) let TN = (1 + A1) (1 + A2) (1 + an), find the general term formula of TN and sequence {an}

(I) prove: from the known, we get an + 1 = an2 + 2An, ∵ an + 1 + 1 = (an + 1) 2. ∵ A1 = 2, ∵ an + 1 ＞ 1. Take logarithm on both sides, we get LG (an + 1 + 1) = 2lg (an + 1), that is LG (an + 1 + 1) LG (an + 1) = 2. The sequence {LG (1 + an)} is an equal ratio sequence with Lg3 as the first term and the common ratio of 2. (II) from (I), we get LG (an + 1) = 2n − 1lg3 = lg32n − 1, ∵ an + 1 = 32n − 1, ∵ an = 32n − 1 ∴Tn=（1+a1）（1+a2）（1+an）=3×321×322××32n−1=31+2+22++2n−1=32n−1．

In the known sequence an, A1 = 2, a (n + 1) = an ^ 2 + 2 * an. 1, prove; LG (1 + an) is an equal ratio sequence
In the known sequence an, A1 = 2, a (n + 1) = an ^ 2 + 2 * an. 1, prove; LG (1 + an) is an equal ratio sequence
2. Let TN = (1 + A1) (1 + A2)... (1 + an), find TN

1.
∵a（n+1）=（an）^2+2an
∴a(n+1)+1=(an+1)^2.(1)
And ∵ A1 = 2 > 1
Yi Zhian > 0
If the common logarithm is taken for both sides of (1), then:
lg[a(n+1)+1]=2lg(an+1)
And ∵ an + 1 ≠ 1
∴lg[a(n+1)+1]/lg(an+1) = 2
That is to say, the sequence {LG (an + 1)} is an equal ratio sequence with the common ratio of 2 and the first term of Lg3
two
lg(an+1)=lg3 * (2^n -1)
lgTn=lg(1+a1)+lg(1+a2)+...+lg(1+an)
=lg3*[2-1+2^2-1+...+2^n-1]
=lg3*[2^(n+1)-n-2]
∴Tn=3^[2^(n+1)-n-2]