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
You use the FOIL method. First terms Outer terms Inner terms Last terms.
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.
What is the sum of the first 27 terms of the geometric sequence -3, 3, - 3, 3, . . . ?
no it is a binomial. terms in an algebriac expression are separated by addition or subtraction ( + or -) symbols and must not be like terms. then just count the terms. one term = monomial, 2 terms = binomial, 3 terms = trinomial. More than 3 terms are usually just referred to as polynomials.
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
4, 8, 12
it is 8.
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