The number whose square equals 2.25 is () The number whose square equals 2.25 is（

The number whose square equals 2.25 is () The number whose square equals 2.25 is（

one point five

How to calculate the square of the sum or subtraction of two numbers?
For example, (x + 6) ^ 2, (X-6) ^ 2 specific point,

Formula: (a + b) ^ 2 = a ^ 2 + 2Ab + B ^ 2
(x+6)^2
= x^2+2*x*6+6^2
=x^2+12x+36
Formula: (a-b) ^ 2 = a ^ 2-2ab + B ^ 2
(x－6)^2
=x^2-2*x*6+6^2
=x^2-12x+36

Who knows the calculation method of adding the squares of two non adjacent numbers?
The sum of the squares of two nonadjacent numbers=
Multiplication of the squares of two adjacent numbers=
The square division of two adjacent numbers=
The cube subtraction of two adjacent numbers=
The cubic addition of two adjacent numbers=
The cube multiplication of two adjacent numbers=
The cubic division of two adjacent numbers=
That's right. That's right. I'll raise the reward

Sum of squares of two non adjacent numbers = m ^ 2 + n ^ 2 multiplication of squares of two adjacent numbers = n ^ 2 * (n + 1) ^ 2 = [n (n-1)] ^ 2 division of squares of two adjacent numbers = n ^ 2 / (n + 1) ^ 2 = [n / (n-1)] ^ 2 subtraction of cubes of two adjacent numbers = n ^ 3 - (n + 1) ^ 3 = [n - (n + 1)] [n ^ 2 + n (n + 1) + (n + 1) ^ 2

A1 = 3, a (n + 1) = 3an-2, write its first five terms, and find the general term formula
A1 = 3 = = = > 1 is subscript
A (n + 1) = 3an-2 = = = > (n + 1), n is the subscript

a1=2*3^(1-1)+1=3
a2=3a1-2=2*3^(2-1)+1=7
a3=3a2-2=2*3^(3-1)+1=19
a4=3a3-2=2*3^(4-1)+1=55
a5=3a4-2=2*3^(5-1)+1=163
.
.
.
an=3^n-[3^（n-1）-1]
=3*3^n-1-3^（n-1）+1
=2*3^n-1+1

In known sequence an, A1 = 1, a (n + 1) = 3an + 2 ^ n, find the general term formula an

Given that the sum of the first n terms of sequence {an} is: SN = 2n & sup2; + N + 1, find the general term formula of sequence {an}

Sn=2n2+n+1
When n > 1
a(n)=S(n)-S(n-1)=2n2+n+1-[2(n-1)2+n]=2n2+n+1-[2n2-3n+2]=4n-1
When n = 1
a(1)=S(1)=4
So a (n)=
4,n=1
4n-1,n>=2.

Given the first n terms of sequence and Sn = 3 ^ n + 1, find its general term formula

When n ≥ 2
an=Sn-S(n-1)
=3^n+1-3^(n-1)-1
=3*3^(n-1)-3^(n-1)
=2*3^(n-1)
When n = 1
a1=S1=4
Does not conform to an = 2 * 3 ^ (n-1)
therefore
n=1,an=4
n≥2,an=2*3^(n-1)

The first few terms of sequence {a} and Sn = 3 ^ n + 1, find its general term formula

n≥2
Then s (n-1) = 3 ^ (n-1) + 1
So an = SN-S (n-1) = 2 * 3 ^ (n-1)
a1=S1
So A1 = 3 + 1 = 4
An2 * 3 ^ (n-1) when n ≥ 2 is not met
therefore
an=
4,n=1
2*3^(n-1),n≥2

Given the first n terms of sequence {an} and Sn = N2 + N, then its general term formula is an = ()
A. nB. 2nC. 2n+1D. n+1

A1 = S1 = 1 + 1 = 2, an = sn-sn-1 = (N2 + n) - [(n-1) 2 + (n-1)] = 2n. When n = 1, 2n = 2 = A1, an = 2n

Let the sum of the first n terms of the sequence an be Sn, satisfy Sn = 5N ^ 2 + 3N + 1, find the general term formula of the sequence an and judge whether an is an arithmetic sequence

∵ Sn = 5N & # 178; + 3N + 1, then sn-1 = 5 (n-1) &# 178; + 3 (n-1) + 1 = 5N & # 178; - 7n + 3
When n ≥ 2, an = sn-sn-1 = 10n-2
When n = 1, A1 = S1 = 9
When n = 1, an = 10n-2 does not hold for an, so an is not an arithmetic sequence
Hope to help you,:)