Use the matchstick to make the figure as follows: No.1: 4 NO.2: 12 NO.3: 24 NO.4: 40 No.5: 60 no.n: help

Use the matchstick to make the figure as follows: No.1: 4 NO.2: 12 NO.3: 24 NO.4: 40 No.5: 60 no.n: help

N*(2N+2)

Use 8 matchsticks to make a square first, and then use 10 matchsticks to make a figure equal to its area
Use 8 matchsticks to make a square first. (I can't send the picture now, sorry, but I should be able to imagine that there are 2 matchsticks on each side). Then use 10 matchsticks to make a figure equal to its area. Finally, add 2 matchsticks and move 1 matchstick to reduce the area of this figure by a quarter
Please draw a diagram:
(summer homework for grade three of primary school)
Helpless ah, the third grade problem, I read for half an hour will not, shame ah~~~~

Look at the picture

Given the sequence an satisfies an > 0, Sn = [(an + 1) / 2] ^ 2, BN = (- 1) ^ n * Sn, find B1 + B2 + +bn

4sn = (an + 1) ^ 24s (n-1) = [a (n-1) + 1] ^ 2An = SN-S (n-1) so subtracting 4An = (an + 1) ^ 2 - [a (n-1) + 1] ^ 2 (an + 1) ^ 2-4an = [a (n-1) + 1] ^ 2 (an-1) ^ 2 = [a (n-1) + 1] ^ 2an-1 = a (n-1) + 1 or an-1 = - A (n-1) - 1An = a (n-1) + 2 or an = - A (n-1) an > 0, so an = - A (n-1) does not hold

It is known that each item of the equal ratio sequence {an} is a positive number, and the sequence {BN} satisfies BN = log2an, and B1 + B2 + B3 = 3, b1b2b3 = - 3 to find the general term an=

Because BN = log2an, then BN is an arithmetic sequence. If BN tolerance is D, then B1 + B2 + B3 = 3 deduces 3b1 + 3D = 3, and then d = 1-b1. Then B1 ^ 3 + 3 * D * B1 ^ 2 + 2 * d ^ 2 * B1 = - 3 can be deduced from the question: b1b2b3 = - 3, then B1 = - 1 or B1 = 3 can be solved if B1 = - 1D = 1-b1 = 2, B2 = B1 + D = 1; A1 = 0.5, A2 = 2

Let {an} be a sequence of arithmetic numbers and {BN} be a sequence of proportional numbers with positive items, and A1 = B1 = 1, A2 + B3 = A3 + B2 = 7 (1) find the general formula of {an}, {BN}
Let {an} be a sequence of arithmetic numbers and {BN} be a sequence of proportional numbers with positive terms, and A1 = B1 = 1, A2 + B3 = A3 + B2 = 7 (1) find the general term formula of {an}, {BN} and (2) find the first n term and Sn of {an / BN}

The tolerance D, common ratio q is substituted into: D = q = 2An = 2N-1; BN = 2 ^ (n-1) an / BN = (2n-1) / 2 ^ (n-1) Sn = 1 + 3 / 2 + 5 / 2 ^ 2 + +(2n-1)/2^(n-1)Sn/2=1/2+3/2^2+5/2^3+…… +(2n-3) / 2 ^ (n-1) + (2n-1) / 2 ^ n, right dislocation: SN = 6 - (2n + 3) / 2 ^ (n-1)

Given the first term A1 > 1, common ratio Q > 0, let BN = log2an, (n belongs to n *) and B1 + B2 + B3 = 6, b1b2b3 = 0. (1) find the general term formula of an
(2) Let the first n terms of BN and Sn, when S1 / 1 + S2 / 2 +. + Sn / N is the maximum, the value of n is obtained

(1) : BN = log2an, an = 2 ^ (BN) B1 = log (2) A1A2 = A1 * q, B2 = log (2) a1 + log (2) QA3 = A1 * q ^ 2, B3 = log (2) a1 + 2log (2) Q. an = A1 * q ^ (n-1) BN = log (2) a1 + (n-1) log (2) Q, so BN is the common ratio of arithmetic sequence, log (2) Q and b1b2b3 = 0, so there must be at least one term here

Let {an} be the sequence of equal proportion numbers with positive items, BN = iog2an, if B1 + B2 + B3 = 3, b1b2b3 = - 3, find the general formula of the sequence of equal proportion numbers
Master who can solve this problem, thank you on your knees

An is an equal ratio sequence
Since BN = log2an, then BN is an arithmetic sequence, let BN tolerance be d
Then B1 + B2 + B3 = 3 deduces 3b1 + 3D = 3 and then d = 1-b1
Then from the question: b1b2b3 = - 3, we can deduce B1 ^ 3 + 3 * D * B1 ^ 2 + 2 * d ^ 2 * B1 = - 3
So we can get B1 = - 1 or B1 = 3
If B1 = - 1
d=1-b1=2,b2=b1+d=1；
a1=0.5,a2=2；
So the common ratio is 4
an=0.5*4^n;
If B1 = 3
d=1-b1=-2,b2=b1+d=1
a1=8,a2=2;
So the common ratio is 0.25;
an=8*(0.25)^n
explain:
It is said in the question that all the items of an are positive, then the common ratio is positive, in order to ensure that log2q is meaningful

The general term formula of known sequence {an} is an = n, let BN = an / 2 ^ n, and prove: B1 + B2 +. + BN

Let the sum of n terms of BN be Sn, then Sn = 1 / 2 + 2 / 2 ^ 2 + 3 / 2 ^ 3 +. + n / 2 ^ n 2Sn = 1 + 2 / 2 ^ 1 + 3 / 2 ^ 2 + 4 / 2 ^ 3 +. + n / 2 ^ (n-1), so 2Sn Sn = 1 + (2 / 2 ^ 1-1 / 2 ^ 1) + (3 / 2 ^ 2-2 / 2 ^ 2) +... + [n / 2 ^ (n-1) - (n-1) / 2 ^ (n-1)] - N / 2 ^ n Sn = 1 + 1 / 2

The known sequence {an} satisfies: A1 = 1, and an-a (n-1) = 2n. Find A2, A3, A4. Find the general term an of the sequence {an}

a2-a1=2*2=4a2=4+1=5a3-a2=2*3=6a3=6+5=11a4-a3=2*4=8a4=11+8=192、an-a(n-1)=2na(n-1)-a(n-2)=2(n-1)…… A3-a2 = 2 * 3a2-a1 = 2 * 2 add, middle positive and negative offset, an-a1 = 2 * 2 + 2 * 3 + +2n=2(2+3+…… +n)=2*(n+2)(n-1)/2=n²+n-2a1=1...

The general term formula of sequence {an} is an = an-1 + 2n, A1 = 2!

an-a(n-1)=2n
a(n-1)-a(n-2)=2(n-1)
a(n-2)-a(n-3)=2(n-2)
.
a2-a1=2X2=4
Add the above n-1 terms to get: an-a1 = n ^ 2 + n-2
The solution is: an = n ^ 2 + n