Multiply them together.
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 sum of the 1st n terms is : N(3N-1)/2 Explanation : The sum from 1 to N of (3m-2) = 3 * sumFrom1toN(m) - sumFrom1toN(2) = 3 * (N*(N+1)/2) -2*N = N(3N-1)/2 For N=10 => 145
In an arithmetic progression the difference between each term (except the first) and the one before is a constant. In a geometric progression, their ratio is a constant. That is, Arithmetic progression U(n) - U(n-1) = d, where d, the common difference, is a constant and n = 2, 3, 4, ... Equivalently, U(n) = U(n-1) + d = U(1) + (n-1)*d Geometric progression U(n) / U(n-1) = r, where r, the common ratio is a non-zero constant and n = 2, 3, 4, ... Equivalently, U(n) = U(n-1)*r = U(1)*r^(n-1).
Find the Sum to n terms of the series 5 5+55+555+ +n Terms
You.... have to apply this formula! n(n+1)/2 and n is the no. of 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.
RAMANUJANRAMANUJAN
The sum of the 1st n terms is : N(3N-1)/2 Explanation : The sum from 1 to N of (3m-2) = 3 * sumFrom1toN(m) - sumFrom1toN(2) = 3 * (N*(N+1)/2) -2*N = N(3N-1)/2 For N=10 => 145
Hey guys....There is no correct simple general formula for sum to n terms of the series1+1/2+1/3+1/4+ ............. + 1/nThe following expression is relatively a very good approximation.S = ln(n + 0.5) + 0.5772 + 0.03759/(n*n + 1.171)Deviation from the actual value fluctuates but remains relatively low.
The nth term of the series is [ 4/2(n-1) ].
A product of 3 and N would be 3 and N multiplied together, so the product would be 3N. To get a numeric answer, you would first need to find what the value of N is.
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.
In an arithmetic progression the difference between each term (except the first) and the one before is a constant. In a geometric progression, their ratio is a constant. That is, Arithmetic progression U(n) - U(n-1) = d, where d, the common difference, is a constant and n = 2, 3, 4, ... Equivalently, U(n) = U(n-1) + d = U(1) + (n-1)*d Geometric progression U(n) / U(n-1) = r, where r, the common ratio is a non-zero constant and n = 2, 3, 4, ... Equivalently, U(n) = U(n-1)*r = U(1)*r^(n-1).
Find the Sum to n terms of the series 5 5+55+555+ +n Terms
You.... have to apply this formula! n(n+1)/2 and n is the no. of terms
The product of any nonzero real number and its reciprocal is the number 1. This can be mathematically given as n multiplied by 1/n, where n represents the nonzero real number. The product of these two terms is 1.
never less than n