1 + 2 + 3 +... What is 365 or tell me the formula

1 + 2 + 3 +... What is 365 or tell me the formula

1+2+3+...365
=1/2×365×（365+1）
=66795
Formula: 1 / 2 * n * (n + 1)

Given that f (1) = 0, AF (n) = BF (n-1) - 1, n is greater than or equal to 2, a 〉 0, B 〉 0, find the expression of F (n)

af(n)=bf(n-1)-1,…………… （1）
Then AF (n-1) = BF (n-2) - 1 （2）
(1) - (2)
a[f(n)-f(n-1)]=b[f(n-1)-f(n-2)]
So [f (n) - f (n-1)] / [f (n-1) - f (n-2)] = B / A
So the sequence {f (n) - f (n-1)} is an equal ratio sequence whose common ratio is B / A,
And f (2) = BF (1) - 1 = - 1
So f (n) - f (n-1) = [f (2) - f (1)] * (B / a) ^ (n-1) = - (B / a) ^ (n-1), n is greater than or equal to 2
That is, f (n) = f (n-1) - (B / a) ^ (n-1) = [AF (n) + 1] / B - (B / a) ^ (n-1)
So f (n) = [a ^ (n-1) - B ^ n] / {[a ^ (n-1)] * (B-A)}, n is greater than or equal to 2
Then when n = 1, f (n) = (1-B) / (B-A) is not equal to 1
So the expression of F (n) is
When n = 1, f (n) = 1;
When n > 1, f (n) = [a ^ (n-1) - B ^ n] / {[a ^ (n-1)] * (B-A)}

(1 / 2) let a > 0, | X|

Because (1 + x) ^ α≈ 1 + α x
So (a ^ n + x) ^ (1 / N) = a [1 + (x / A ^ n)] ^ (1 / N) ≈ a [1 + (1 / N) (x / A ^ n)] = a [1 + X / (Na ^ n)] = a + X / [Na ^ (n-1)]
1000^(1/10)=(1024-24)^(1/10)=(2^10-24)^(1/10)≈2-24/[10*2^9]=2-24/5120=1.9953125

If it is required to express the relationship that a is not equal to 0 in a pair of parentheses after if, the expression that can correctly express the relationship is a) A0 b) a C) a = 0 d) a

The answer is d
Because the implicit meaning of the parentheses after if is to continue to execute the following program when the program in the parentheses is "true", that is, 1. That is to say, if the program is written as if (a), it can be interpreted as executing the following program when a is not equal to 0. In addition, it should be declared that in programming, the conditional judgment value is not 1, that is, 0, and other numbers are equivalent to 1
The expression of a option does not exist, and BC option is just the opposite
This is my answer. I hope I can help you

1^2+2^2+3^2+… +(n-1)^2+n^2=?

n(n+1)(2n+1)/6
Proof of mathematical induction

1! + 2! + 3! ··· n! Formula
It's urgent

In my opinion, there should be no formula for this, only hard calculation, because exclamation operator (factorial) does not have a simple calculation formula. If you have a calculation formula, can you get the exclamation calculation formula by subtracting n-1 from item n,

Why is the general formula divided into n = 1 and N greater than or equal to 2

Because the formulas of some general term formulas are different when n = 1 and N > = 2, that is, the first term of the sequence does not satisfy the general term formula

Seek 1 + 2 + 3 + 4 The formula of n-square and its application

Because: 1 + 2 + 2 + + n = 1 / 2n (n + 1) because: 1 + 2 + 2 + + n = 1 / 2n (n + 1), then (n + 1) * (n + 1) (n + 1) (n + 1) (n + 1) as (n + 1 + 1) (n + 1 + N + 1) (n + 1 + N + 1) (n + 1 + 1) is (n + 1 + 1) as (n + 1) * (n + 1 + 1) (n + 1) * (n + 1 (n + 1) - n * n * n * n * n * n * n * n * n * n * n - (N-N - (n-1 - (N-N-N - (n-n-1 * (n + 1) * (n + 1) (n + 1) (n + 1) (n + 1) * (n + 1) (n + 1) (n + 1) (n + 1 (n + 1) (n + 1) * (n + 1) (n + 1) (n + 1) (n + 1 (n + 1) (n + 1 + + +n*n=1/6n(n+1)(2n+1)

X Cubic - 1... How to decompose the cubic formula
x^2-1=?
X Cubic - 1 =?

x^2-1=(x-1)(x+1)
x^3-1=(x-1)(x^2+x+1)

2006×2007+2007×2008+2006×2008−21+2+3+… +2006+2007．

2006×2007+2007×2008+2006×2008−21+2+3+… +2006+2007=2006×(2008−1)+(2006+1)×2008+2006×2008−2(1+2007)×2007÷2=2006×2008−2006+2006×2008+2008+2006×2008−22008×2007÷2=2006×2008×32008×2007...