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Hey guys....

There is no correct simple general formula for sum to n terms of the series

1+1/2+1/3+1/4+ ............. + 1/n

The 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.

Q: How do you find the sum to n terms of a harmonic progression?

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The sum of the first forty positive integers can be calculated using the formula for the sum of an arithmetic series, which is (n/2)(first term + last term) where n is the number of terms. In this case, the sum is (40/2)(1 + 40) = 820.

An Ashanti number is a concept in mathematics related to the Fibonacci sequence, where each term is the sum of the two preceding ones. Ashanti numbers are formed by starting with two initial values and then generating subsequent terms based on their sum. This concept can be extended beyond Fibonacci numbers to include other sequences derived in a similar manner.

To find the sum of all elements in a matrix in C, you can use nested loops to iterate through each element and accumulate the sum. Here is a basic example: #include <stdio.h> #define ROWS 3 #define COLS 3 int main() { int matrix[ROWS][COLS] = {{1, 2, 3}, {4, 5, 6}, {7, 8, 9}}; int sum = 0; for(int i = 0; i < ROWS; i++) { for(int j = 0; j < COLS; j++) { sum += matrix[i][j]; } } printf("Sum of all elements in the matrix: %d\n", sum); return 0; }

#include <stdio.h> #include <conio.h> void main() { int d[3][3] = { 1, 2, 6, 3, 8, 5, 5, 6, 7 }; int k = 0, j = 0; int sum1 = 0, sum2 = 0; for (j = 0; j < 3; j++) { for (k = 0; k < 3; k++) printf(" %3d", d[j][k]); printf("\n"); } for (j = 0; j < 3; j++) { sum1 = sum1 + d[j][j]; } k = 3 - 1; for (j = 0; j < 3; j++) { if (k >= 0) { sum2 = sum2 + d[j][k]; k--; } } printf("Sum of First diagonal= %d\n", sum1); printf("Sum of Second diagonal= %d", sum2); getch();

first you multiply 33333333334 by 33333333334 (if its kinda hard to count these, they both have ten 3's, and one 4) and then you should get this: 1111111111155555555556 (for this one, there are eleven 1's, ten 5's, and one 6) then you add all these digits together, and should get this: 67 ***p.s. the reason why it's not 1156 is because, in the order of operations, you have to solve the exponents (to the second power) before you can start adding***

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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.

An arithmetic series is the sum of the terms in an arithmetic progression.

RAMANUJANRAMANUJAN

For an Arithmetic Progression, Sum = 15[a + 7d].{a = first term and d = common difference} For a Geometric Progression, Sum = a[1-r^15]/(r-1).{r = common ratio }.

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.

This question is impossible to answer without knowing the difference between successive terms of the progression.

This series is known as the Harmonic Series and it diverges but very, very slowly. For example, the first 100 terms sum to 5.187...., the first 1000 terms to 7.486...., and the first 1000000 terms to 14.392.... There are many proofs of the divergence of this series and an internet search of Harmonic Series will no doubt find many of them.

The sum to infinity of a geometric series is given by the formula Sā=a1/(1-r), where a1 is the first term in the series and r is found by dividing any term by the term immediately before it.

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

yup

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