You.... have to apply this formula! n(n+1)/2 and n is the no. of terms
a+a*r+a*r^2+...+a*r^n a = first number r = ratio n = "number of terms"-1
It is not possible to answer this question without information on whether the terms are of an arithmetic or geometric (or other) progression, and what the starting term is.
The simplest formula is: t(n) = 2n + 5 for n = 1, 2, 3, ... However, that is not the only formula; there are infinitely many polynomial formulae that can be found that give those five terms first, but for the 6th or further terms vary.
Assuming each term is 3 MORE than the previous term t(n) = -13 + 3*n where n = 1, 2, 3, ...
Permutation Formula A formula for the number of possible permutations of k objects from a set of n. This is usually written nPk . Formula:Example:How many ways can 4 students from a group of 15 be lined up for a photograph? Answer: There are 15P4 possible permutations of 4 students from a group of 15. different lineups
You.... have to apply this formula! n(n+1)/2 and n is the no. of terms
RAMANUJANRAMANUJAN
a+a*r+a*r^2+...+a*r^n a = first number r = ratio n = "number of terms"-1
t(1) = 1t(2) = 1t(n) = t(n-2) + t(n-1) for n = 3, 4, 5, ...that is, the first and second terms are 1. After that, each term is the sum of the previous two terms.
It is not possible to answer this question without information on whether the terms are of an arithmetic or geometric (or other) progression, and what the starting term is.
The formula to find the sum of a geometric sequence is adding a + ar + ar2 + ar3 + ar4. The sum, to n terms, is given byS(n) = a*(1 - r^n)/(1 - r) or, equivalently, a*(r^n - 1)/(r - 1)
The differences between terms are 7,9,11,15 The differences of these differences are 2,2,2,2 Thus the formula for the sequence begins with n2 The sequence of numbers minus n2 is 5, 9, 13, 17, 21 The difference between the terms is 4 So the formula continues n2+4n The sequence of numbers minus n2+4n is 1, 1, 1, 1, 1 So the formula for the nth term is n2+4n+1
The simplest formula is: t(n) = 2n + 5 for n = 1, 2, 3, ... However, that is not the only formula; there are infinitely many polynomial formulae that can be found that give those five terms first, but for the 6th or further terms vary.
Assuming each term is 3 MORE than the previous term t(n) = -13 + 3*n where n = 1, 2, 3, ...
Unless otherwise stated any series/sequence starts with n = 1 Then the first three terms are when n = 1, n= 2 and n = 3 and the series formula 4n - 2 gives the related results, 2, 6 and 10.
The series given is an arithmetic progression consisting of 5 terms with a common difference of 5 and first term 5 → sum{n} = (n/2)(2×5 + (n - 1)×5) = n(5n + 5)/2 = 5n(n + 1)/2 As no terms have been given beyond the 5th term, and the series is not stated to be an arithmetic progression, the above formula only holds for n = 1, 2, ..., 5.