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
The product of (n - 8)(n + 2) can be found using the distributive property (also known as the FOIL method for binomials). Multiplying the terms gives: n² + 2n - 8n - 16. Combining like terms results in the expression n² - 6n - 16.
Find the Sum to n terms of the series 5 5+55+555+ +n Terms
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).
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
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 formula for the sum of the first n terms of an arithmetic progression is Sn = n/2 * (a + l), where Sn is the sum, n is the number of terms, a is the first term, and l is the last term.
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
To find the sum of the first 20 terms of an arithmetic progression (AP), we need to first determine the common difference (d) between the terms. Given that the 6th term is 35 and the 13th term is 70, we can calculate d by subtracting the 6th term from the 13th term and dividing by the number of terms between them: (70 - 35) / (13 - 6) = 5. The formula to find the sum of the first n terms of an AP is Sn = n/2 [2a + (n-1)d], where a is the first term. Plugging in the values for a (the 1st term), d (common difference), and n (20 terms), we can calculate the sum of the first 20 terms.
The product of (n - 8)(n + 2) can be found using the distributive property (also known as the FOIL method for binomials). Multiplying the terms gives: n² + 2n - 8n - 16. Combining like terms results in the expression n² - 6n - 16.
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 nth term of the series is [ 4/2(n-1) ].
Find the Sum to n terms of the series 5 5+55+555+ +n Terms
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).