1,5,21,85341

1,5,21,85341

The first number multiplied by 4 and 1 is the last one

It is known that SN is the sum of the first n terms of the arithmetic sequence an (n ∈ n *). If S7 > S5, then ()
A. a6+a7＜0B. S9＞S3C. a7+a8＞0D. S10＞S4

∵ S7 ＞ S5, ∵ A6 + A7 ＞ 0, ∵ {an} is the arithmetic sequence, ∵ A4 + A9 = A5 + A8 = A6 + A7 ＞ 0, ∵ A4 + A5 + A6 + A7 + A8 + A9 ＞ 0 ∵ s9-s3 ＞ 0, ∵ S9 ＞ S3

Let the sum of the first n terms of the arithmetic sequence {an} be Sn, and A3 + A5 + A7 = 15, then S9 = ()
A. 18B. 36C. 45D. 60

From A3 + A5 + A7 = 3A5 = 15, the solution is A5 = 5, then S9 = 9 (a1 + A9) & nbsp; 2 = 9A5 = 45

It is proved that the square difference of two consecutive odd numbers is a multiple of 8
Tip: let two consecutive odd numbers be 2A + 1, 2A + 3, (a is a positive integer)

Let two consecutive odd numbers be 2A + 1,2a + 3, (a is a positive integer)
Using the square difference formula: (2a + 1 + 2A + 3) times (2a + 3-2a-1) = (4a + 4) times 2 = 4 times (a + 1) times 2
It must be a multiple of 2 times 4 = 8
Your adoption is the driving force of my answer!

Is the sentence "the square difference of two consecutive odd numbers must be a multiple of 8"? Give reasons

The reason is that: (2n + 1) 2 - (2n-1) 2 = [(2n + 1) + (2n-1)] [(2n + 1) - (2n-1)] = (2n + 1 + 2n-1) (2n + 1-2n + 1) = 4N × 2 = 8N, 8N △ 8 = n

When n is a positive integer, the square difference of two consecutive odd numbers must be a multiple of 8

It is proved that when n is a positive integer, then two consecutive odd numbers are 2N-1 and 2n + 1 respectively
∴ (2n＋1)^2-(2n-1)^2
＝(2n＋1＋2n-1)(2n＋1-2n＋1)
＝4n×2
＝8n
Because the above formula contains a factor of 8, and N is a positive integer
So 8N is divisible by 8
The square difference of these two consecutive odd numbers is a multiple of 8

Try to explain that the square difference of two consecutive odd numbers must be odd

Let a be an odd number
a,a+2
(a + 2) ^ 2-A ^ 2 = 4A + 4 is even

Proof: when n is a number, the square difference (2n + 1) ^ - (2n-1) ^ of two consecutive odd numbers is a multiple of 8

It is proved that two consecutive odd numbers are 2n ± 1
The square difference = (2n + 1) & sup2; - (2n-1) & sup2;
=（2n+1+2n-1)+(2n+1-2n+1)
=4n×2
=8N is a multiple of 8

Is the sentence "the square difference of two consecutive odd numbers must be a multiple of 8"? Give reasons

The reason is that: (2n + 1) 2 - (2n-1) 2 = [(2n + 1) + (2n-1)] [(2n + 1) - (2n-1)] = (2n + 1 + 2n-1) (2n + 1-2n + 1) = 4N × 2 = 8N, 8N △ 8 = n

Try to explain: the product of four consecutive odd numbers minus the difference of 1 can be divided by 8. It is known that the trilateral lengths a, B and C of △ ABC satisfy the following conditions: the square of a + the square of B + the square of c-a
1. Try to explain: the product of four consecutive odd numbers minus the difference of 1 can be divided by 8
2. Given that the lengths of three sides a, B and C of △ ABC satisfy a's square + B's square + C's Square - AB AC BC = 0, try to judge the shape of triangle ABC

1. Let the four odd numbers be 2k-3, 2k-1, 2K + 1, 2K + 3 (k is not less than 2). The product minus 1 is reduced to 16K ^ 4-40k ^ 2 + 8, and each term can be divisible by 8, so the whole formula can also be divisible by 8. 2. A ^ 2 + B ^ 2 + C ^ 2 = AB + AC + BC two sides multiplied by 2 is reduced to (a ^ 2-2ab + B ^ 2) + (a ^ 2-2ac + C ^ 2) + (b ^ 2)