How to make 4 triangles with 7 matchsticks?

It's a three-dimensional one. First place three of them into a triangle, and then connect the other three with the vertices of the first triangle

The number of triangles that can be placed into different triangles is______ One

According to the perimeter of 7 and the trilateral relationship of the triangle, there are only two different kinds of triangles, the length of which is 2, 2, 3 or 3, 3, 1. Other combinations can not satisfy the trilateral relationship of the triangle

Using a small wooden stick to make a figure, one uses six, two uses 11, three uses 16, ten uses () and () uses 401. The rule is ()

51, 80, x, y, 5x + 1 = y,

How many sticks do you need? How many sticks do you need

The formula is 5N + 1, n is the number, 9 is 46, 2006 is 401

bn=(a1+2a2+3a3+4a4+…… +nan)/(1+2+3+4+…… +n) A necessary and sufficient condition to prove that an is an arithmetic sequence and BN is an arithmetic sequence

It is proved that if an is an arithmetic sequence, then BN is an arithmetic sequence. (sufficiency) let Sn = a1 + 2A2 + 3a3 + 4a4 + +Nan = (a1 + A2 +... + an) + (A2 + a3 +... + an) +... + (a (n-1) + an) + an = half of the sum of the first n terms of N ans + half of the sum of the first n terms of 1 an = n (n + 1) (a1 + an) / 2, then BN = a1 + an = 2A1 +

Let BN satisfy: B1 = 1 / 2, BN + 1 = BN ^ 2 + bn1) prove: BN + 1 / 1 = BN / 1-bn + 1 / 12) if TN = B1 + 1 / 1 + B2 + 1 / 1 +. + BN + 1 / 1, find the minimum value of TN
Just solve the second question, TN = B1 + 1 / 1 + B2 + 1 / 1 +... + BN + 1 / 1, and find the minimum value of TN

　　（1）
Because B (n + 1) = BN ^ 2 + BN
The reciprocal is: 1 / b (n + 1) = 1 / BN - 1 / BN + 1
The above formula is transformed into 1 / BN + 1 = 1 / bn-1 / b (n + 1)
tn=b1+1/1+b2+1/1+.+bn+1/1=1/b1-1/b2+1/b2-1/b3+………… +1/bn-1/b(n+1)
=1/b1-1/b(n+1)
Because, BN + 1 = BN ^ 2 + BN, using the knowledge of function. Y = BN ^ 2 + BN, it is easy to prove that y is (1 / 2, + infinity)
B (n + 1) is an increasing sequence, while 1 / b (n + 1) is a decreasing sequence, and - 1 / b (n + 1) is an increasing sequence
When n approaches infinity, limb (n + 1) = limbn ^ 2 + BN, the limit of BN is
1 / b1-1 / b (n + 1) = 2-1 / b (n + 1) infinitely approaches to 2, but it can never be achieved
When n = 1, the minimum value is 2 / 3
My answer is the most correct,

Mathematics problem: known sequence {BN} is arithmetic sequence, B1 = 1, B1 + B2 +... + B10 = 145
The known sequence {BN} is an arithmetic sequence, B1 = 1, B1 + B2 +... + B10 = 145
Let an = loga (1 + 1 / BN) be the general term of the sequence {an}, where a is greater than 0 and a is not equal to 1. Note that SN is the sum of the first n terms of the sequence {an}. Compare Sn with 1 / 3 loga BN + 1 and prove your conclusion
It's 1 / 3 times

(1)Bn=3n-2
b1+b2+b3+.+b10=10b1+d+2d+.+9d
=10+45d=145
Then d = 3
Because BN = B1 + (n-1) * D
So BN = 3n-2
(2) The question is not clear enough. Is it the product of one-third times logabn or 1 divided by 3 times logabn?
Refer to online answers:
Let the tolerance of the sequence {BN} be d. from the meaning of the problem, we can get that 〈 BN = 3n-2
(2) It is proved that we know from BN = 3n-2
Sn=loga(1+1)+loga(1+ )+… +loga(1+ )
=loga〔(1+1)(1+ )… (1+ )〕
And logabn + 1 = loga (1 +) and the size of
Let n = 1, have (1 + 1)=
Let n = 2, have (1 + 1) (1)+
Conjecture: (1 + 1) (1 +) (1+ )＞ (*)
① When n = 1, it has been proved that (*) is true
② Suppose n = K (K ≥ 1), then (*) holds, that is, (1 + 1) (1 +) (1+ )＞
Then when n = K + 1,
That is, when n = K + 1, (*) holds
It is known from (1) and (2) that the expression (*) holds for any positive integer n
So, when a > 1, Sn > logabn + 1; when 0 < a < 1, Sn < logabn + 1

The sequence an = (1 / 2) ^ n, the sequence {BN} satisfies BN = 3 + log4an, let TN = | B1 | + | B2 | +... + | BN |, find TN

bn=3+log4(1/2)^n=3+log42^(-n)=3-n/2
bn=3-n/2=6
S1=b6+b7+b8+.+bn=(b6+bn)*(n-5)/2= -n^2/2+11n/2-15
S2=(b1+b5)*5/2=15/2
Tn=S2-S1=n^2/2-11n/2+45/2

Given that the sequence an is an arithmetic sequence, (BN) is an equal ratio number, and A1 = B1 = 2, B4 = 54, a1 + A2 + a3 = B2 + B3, find (1) the general term formula of the sequence (BN), and (2) the sum S10 of the first 10 terms of the sequence (an)

(1) Because {BN} is an equal ratio sequence, B1 = 1, B4 = 54, so 2q ^ 3 = 54, q = 3, so BN = 2 * 3 ^ n-1
(2) From (1), B2 = 6, B3 = 18, so a1 + A2 + a3 = 3A1 + 3D = 24, so d = 6, so an = 6n-4, so the first n terms of an and Sn = n (3n-1), so S10 = 290
I don't know how to ask~

Sequence an is the arithmetic sequence BN is the proportional sequence A1 = B1 = 3 A2 = B2 A3 ratio B3 = 5 to 9 to find the general term formula

It can be seen from the meaning of the title
a1+d=b1q
(a1+2d)/b1q²=5/9
By introducing A1 = B1 = 3 into the above equations, we can get the following results:
3+d=3q
(3+2d)/3q²=5/9
The results are as follows
D = - 6 / 5, q = 3 / 5 or D = 6, q = 3
Then when d = - 6 / 5, q = 3 / 5, an = 21 / 5-6N / 5 BN = 3 (3 / 5) ^ n-1
When d = 6, q = 3, an = - 3 + 6N BN = 3 ^ n (the nth power of 3)

How to make 4 triangles with 7 matchsticks?