1. Sequence: 1, - 2,3, - 4 What is the sum of the first 99 items of (- 1) ^ (n-1)? 2. Equal ratio sequence 1, A1, A2, A3 Then A1 * A2 * A3 * *a2n=?

1. Sequence: 1, - 2,3, - 4 What is the sum of the first 99 items of (- 1) ^ (n-1)? 2. Equal ratio sequence 1, A1, A2, A3 Then A1 * A2 * A3 * *a2n=?

1. Sequence: 1, - 2 The sum of the first 99 terms of (- 1) ^ (n-1) = (1-2) + (3-4) +... + (97-98) + 99 = - 98 / 2 + 99 = 502 There are 2n + 2 terms A1 * A2 * A3 * *A2N = 1 × Q ^ (1 + 2 +... + 2n) = q ^ 2n (2n + 1) / 2 = q ^ n (2n + 1) 2 = 1 × Q ^ (2n + 1) = q ^ (2n + 1)

The sum of the first n terms of the sequence {(n + 1) 3 ^ n}

The typical "equal difference * equal ratio" sequence is subtracted by dislocation
Sn=2*3^1+3*3^2+4*3^3 +...+(n+1)*3^n (1)
3Sn= 2*3^2+3*3^3+...+ n*3^n + (n+1)*[3^(n+1)] (2)
(1) If you subtract (2), you get
-2Sn= 6+(3^2+3^3+...+3^n) - (n+1)*[3^(n+1)]
(note that there is an equal ratio sequence in the middle, but only n-1 terms)
-2Sn=6 + 3^2[1-3^(n-1)]/(1-3) - (n+1)*[3^(n+1)]
Simplify, simplify
Sn=(3/4) * { [(2n+1) * 3^n] -1 }
(this kind of index calculation is very complicated. Some people can do it in mind. I admire it!)

How to find the general term of sequence a (n + 1) + a (n) = 2 × 3 ^ (n-1)
The middle is + or not - some people say to construct a (n + 1) - A (n) = x [a (n) - A (n-1)], but how to find x?
A (n) is an equal ratio sequence

Because a (n) is equal, let the common ratio be q, so a (n + 1) = Q * a (n)
So a (n + 1) + a (n) = (Q + 1) * a (n)
Because a (n + 1) + a (n) = 2 * 3 ^ (n-1)
So a (n) + a (n-1) = 2 * 3 ^ (n-2) = (Q + 1) * a (n-1)
A (n) / a (n-1) = 3, that is, q = 3
So a (n + 1) + a (n) = (Q + 1) * a (n) = 4A (n)
So a (n) = (3 ^ n-1) / 2
It should be. Let's see if there are any problems

If a n = 1 / [(3 ^ n) - 1], prove: the first n terms and Sn of a n

Let BN = 1 / 2 ^ n
Because when n ≥ 1, 2 ^ n ≤ 3 ^ n-1
So BN ≥ an
Let the sum of BN be TN
Then TN > SN
Tn=b1+…… +bn=1/2*(1-1/2^n)/(1-1/2)=1-2^n