The first three terms for the expression 2n-1 can be found by substituting n with the first three consecutive integers. When n=1, the expression becomes 2(1)-1 = 1. When n=2, the expression becomes 2(2)-1 = 3. When n=3, the expression becomes 2(3)-1 = 5. Therefore, the first three terms are 1, 3, and 5.
29
The first three terms for the expression 2n-6 are obtained by substituting n with consecutive integers. When n=1, the expression evaluates to -4; when n=2, the expression evaluates to -2; and when n=3, the expression evaluates to 0. Therefore, the first three terms are -4, -2, and 0.
You use the FOIL method. First terms Outer terms Inner terms Last terms.
What is the sum of the first 27 terms of the geometric sequence -3, 3, - 3, 3, . . . ?
First off, it is NOT A QUINTIC! Typically a polynomial of four or more terms is called "a polynomial of n terms", where n is the number of terms. Only the one, two, and three term polynomials are referred to by a particular naming convention.
To prove that 2n1 o(2n), we need to show that the growth rate of 2n1 is strictly less than the growth rate of 2n as n approaches infinity. This can be demonstrated by using the formal definition of little-o notation and showing that the limit of (2n1) / 2n as n approaches infinity is equal to 0.
Which sequence? Oh, that one! The first three terms are 1, 2 and 72.
3 7 11
To find the first 5 terms, plug 1, 2, 3, 4 and 5 in for n:3*1-3 = 03*2-3 = 33*3-3 = 63*4-3 = 93*5-3 = 12The first five terms are 0, 3, 6, 9 and 12.
Even numbers : they can be written 2n 2n1+2n2+....+2nm = 2(n1+n2+....+nm) so it's always an even number 2n1x 2n2x....x2nm = 2(n1+n2+....+nm) which is always an even number Odd numbers: they canbe written 2n+1 (2n1+1) + (2n2+1) + ....+ (2nm +1) = 2(n1+n2+....+nm) + m which is even if m is even and odd if m is odd, so it depend of the number of terms (2n1+1) x (2n2+1) x ....x (2nm +1) is always odd cause the last term of the expansion is always +1 and other terms have at least 2 as factor
it is 8.
4, 8, 12
3
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n = 1 ---> 3 * 1 + 3 = 6 n = 2 ---> 3 * 2 + 3 = 9 first 2 have difference of 3, so you can keep adding 3 to find more terms. 6, 9, 12, 15
The first four terms are 3 9 27 81 and 729 is the 6th term.
Suppose the first term is a, the second is a+r and the nth is a+(n-1)r. Then the sum of the first five = 5a + 10r = 85 and the sum of the first six = 6a + 15r = 123 Solving these simultaneous equations, a = 3 and r = 7 So the first four terms are: 3, 10, 17 and 24