Sequence 1 / (1 + 2), 1 / (2's square + 4),. 1 / (n's square + 2n) The sum of the first n terms of

Sequence 1 / (1 + 2), 1 / (2's square + 4),. 1 / (n's square + 2n) The sum of the first n terms of

1/(n^2+2n)）=[1/n - 1/(n+2)]/2
The sum of the first n terms = (1-1 / 3) / 2 + (1 / 2-1 / 4) / 2 +... + [1 / n-1 / (n + 2)] / 2
Split term cancellation = [1 + 1 / 2 - 1 / (n + 1) - 1 / (n + 2)] / 2 = 3 / 4 - (2n + 3) / [2 (n + 1) (n + 2)]

21.5 sq. - 3x21.5 + 1.5 sq. - 100
Simple calculation method

21.5 sq. - 3x21.5 + 1.5 sq. - 100
=(21.5-1.5)²-100
=20²-100
=400-100
=300
Using 21.5 square - 3x21.5 + 1.5 square is a complete square formula to achieve the purpose of simplicity

Sequence 1,4,9, (), 1,0

8
5^0,4^1,3^2,2^3,1^4,0^5

SN is the sum of the first n terms of the sequence {an}, A1 = 2 / 9 and an = Sn · sn-1 (n > = 2)
(1) The general formula for finding an;
(2) When n is the value, Sn is the maximum, and the maximum value is obtained

From the meaning of the question: SN-S (n-1) = an; 1 / sn-1 / S (n-1) = (s (n-1) - Sn) / (SN · s (n-1)) = - an / an = - 1; S1 = A1 = 2 / 9; so the sequence {1 / Sn} is an arithmetic sequence with a tolerance of - 1 and a first term of 9 / 2, so 1 / Sn = 1 / S1 + (n-1) * (- 1) = 11 / 2-N; so Sn = 2 / (11-2n); so an = sn-sn-1 = 4 / [(...)

It is known that the sum of the first n terms of the sequence {an} is Sn, and an = Sn * s (n-1) (n ≥ 2, Sn ≠ 0), A1 = 2 / 9
Prove: {1 / Sn} is arithmetic sequence
Finding the set of natural numbers n satisfying an > A (n-1)
Please feel free to give me advice!

Certification:
Because an = Sn * s (n-1) (n ≥ 2, Sn ≠ 0)
So SN-S (n-1) = Sn * s (n-1) (n ≥ 2, Sn ≠ 0)
Because Sn ≠ 0, so Sn * s (n-1) ≠ 0
Divide both sides of the equation by SN * s (n-1) at the same time
1 / S (n-1) - 1 / Sn = 1, i.e. 1 / Sn = 1 / S (n-1) - 1 (n ≥ 2)
So the sequence {1 / Sn} is an arithmetic sequence with 9 / 2 as the first term and - 1 as the tolerance
Then 1 / Sn = 1 / S1 + (n-1) d = 9 / 2 + (n-1) * - 1 = 11 / 2-n
So Sn = 1 / (11 / 2-N) = 2 / (11-2n), then s (n-1) = 2 / [11-2 (n-1)] = 2 / (13-2n)
So an = SN-S (n-1) = 2 / (11-2n) - 2 / (13-2n) = 4 / [(2n-11) * (2n-13)] (n ≥ 2)
When n = 1, A1 does not conform to an = 4 / [(2n-11) * (2n-13)]
When n = 2, A2 = 4 / 632, an = 4 / [(2n-11) * (2n-13)]
From the square difference formula:
(2n-11)*(2n-13)=[(2n-12)+1]*[(2n-12)-1]=[2(n-6)]^2-1=4*(n-6)^2-1
Then an = 4 / [4 * (n-6) ^ 2-1]
Let f (n) = 4 * (n-6) ^ 2-1
When 20, f (n) decreases in turn
So an = 4 / [4 * (n-6) ^ 2-1] increases in turn, which conforms to an > A (n-1) and A5 = 4 / 3;
When n = 6, A6 = - 4A6;
When n > 7, 4 * (n-6) ^ 2-1 > 0 and f (n) increases in turn, then an decreases in turn, which is not in line with the meaning of the problem;
So to sum up, the set of natural number n satisfying an > A (n-1) is {3,4,5,7}

Sequence {an}, A1 = 4 / 3, A2 = 13 / 9, when n > = 3, an-an-1 = 1 / 3 (an-1-an-2), find the general term formula of {an}, what are the first n terms and Sn

Let BN = an-a (n-1); then B (n-1) = a (n-1) - A (n-2); an-an-1 = 1 / 3 (an-1-an-2), then BN = 1 / 3B (n-1); BN is an equal ratio sequence with a common ratio of 1 / 3; BN = (1 / 3) ^ (n-3) B3; B3 = a3-a2 = 1 / 3 (a2-a1) = 1 / 3 (13 / 9-4 / 3) = 1 / 27; BN = (1 / 3) ^ (n-3) (1 / 27) = (1 / 3) ^ n; an-a (n-1) = (1 /

Equal ratio sequence, a1 + A2 + a3 = 14, the square of a1 + the square of A2 + the square of A3 = 84, find A1, A2, A3

If the common ratio is Q, then a &; = A2 / Q, a &; = A &; -; Q, have
a₂`(1/q+1+q)=14 (1)
a₂²(1/q²+1+q²)=84 (2)
(1) Square on both sides of formula
==> a₂²•[2(1/q+1+q)+(1/q²+1+q²)]=196
==> 28a₂+84=196
==> a₂=4
Substituting (1), the solution is q = 2 or q = 1 / 2
So these three numbers are 2, 4, 8

It is known that the sum of (A1) ^ 2 + (A2) ^ 2 + (A3) ^ 2 +. + (an) ^ 2 is a square number
Why can (A1) ^ 2 + (A2) ^ 2 + (A3) ^ 2 +. + [a (n-1)] ^ 2 = 4K or 2K + 1

Because the square is either odd or even
The square or odd of odd number can be expressed as 2K + 1
The square of an even number is definitely divisible by four

If an = 1's square + 2's square + 3's square + + n's Square, then the sum of the first n terms of the sequence 3 / a1,5 / a2,7 / A3 is?

Square of an = 1 + square of 2 + square of 3 + + square of n = 1 / 6 * n (n + 1) (2n + 1) sequence 3 / a1,5 / a2,7 / A3 The general formula is BN = (2n + 1) / an = 6 / (n (n + 1)) s = 6 / (1 * 2) + 6 / (2 * 3) + 6 / (3 * 4) + +6/(n(n+1)) =6[1/(1*2)+1/(2*3)+1/(3*4)+…… +1/(n(n+1))] ...

If the square of the sequence {a1 + A2 + a3 +... + an = a, then a2011=

sn=a^2
an=sn-s(n-1)=a^2-(a-1)^2=2a-1
a2011=2*2011-1=4021