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What is four times the sum of a number and 3?
4(N-3) is the formula, do the math from there and you get "N-3"
What is the equation for 3 times the sum of a number n and 7 equals 16?
3 (n + 7) = 16
How should seven times the sum of a number n and 3 be written as an algebraic expression?
7(n+3)
What is the product of 6 and the sum of 3 times a number n and 5?
6(3n+5)
What is the nth term and sum of n terms of 6 13 24 39?
The simplest rule that will generate the 4 terms, and the nth term isUn = 2n2 + n + 3Then Sn = Sum for k = 1 to n of (2k2 + k + 3)= 2*sum(k2) + sum(k) + sum(3)= 2*n*(n+1)*(2n+1)/6 + n*(n+1)/2 + 3*n= (2n3 + 3n2 + n)/3 + (n2 + n)/2 + 3n= (4n3 + 9n2 + 23n)/6The simplest rule that will generate the 4 terms, and the nth term isUn = 2n2 + n + 3Then Sn = Sum for k = 1 to n of (2k2 + k + 3)= 2*sum(k2) + sum(k) + sum(3)= 2*n*(n+1)*(2n+1)/6 + n*(n+1)/2 + 3*n= (2n3 + 3n2 + n)/3 + (n2 + n)/2 + 3n= (4n3 + 9n2 + 23n)/6The simplest rule that will generate the 4 terms, and the nth term isUn = 2n2 + n + 3Then Sn = Sum for k = 1 to n of (2k2 + k + 3)= 2*sum(k2) + sum(k) + sum(3)= 2*n*(n+1)*(2n+1)/6 + n*(n+1)/2 + 3*n= (2n3 + 3n2 + n)/3 + (n2 + n)/2 + 3n= (4n3 + 9n2 + 23n)/6The simplest rule that will generate the 4 terms, and the nth term isUn = 2n2 + n + 3Then Sn = Sum for k = 1 to n of (2k2 + k + 3)= 2*sum(k2) + sum(k) + sum(3)= 2*n*(n+1)*(2n+1)/6 + n*(n+1)/2 + 3*n= (2n3 + 3n2 + n)/3 + (n2 + n)/2 + 3n= (4n3 + 9n2 + 23n)/6
How would you write out the sum of twice n and 3 times m?
sum = 2n + 3m
What is four times the sum of a number and 3?
4(N-3) is the formula, do the math from there and you get "N-3"
The sum of 6 times a number n and 4 times a number n is -30 what is the number?
10n = -30 n = -3
What is the average of all the integer's of 13 to 37?
To find the (mean) average, add all the numbers and divide by the number of numbers. The sum of a series of digits (in arithmetic progression, like 13, 14, 15, ... 37) is sum = (first + last) x number_of_digits / 2 So their average is: average = ((first + last) x number_of_digits / 2) / number_of_digits = (first + last) / 2 = average of first and last digits! So the average of the numbers 13, 14, 15, ..., 37 is: average = (13 + 37) / 2 = 50 / 2 = 25 To find the sum of n digits starting with m: Sum = m + (m+1) + ... + (m+n-2) + (m+n-1) Rewrite the sum in reverse order: Sum2 = Sum = (m+n-1) + (m+n-2) + ... + (m+1) + m Add the two sums, term by term: Sum + Sum2 = 2 Sum = (m + (m+n-1)) + (m + (m+n-1)) + ... + (m + (m+n-1)) + (m + (m+n-1)) There are n terms, all (m + (m+n-1)), so: 2 Sum = (m + m+n-1) x n Sum = (m + m+n-1) x n / 2 But n is the number of digits, m is the first number and (m+n-1) is the last, so: Sum = (first + last) x number_of_digits / 2
What is the algebraic expression for the sum of 12 and 3 times a number?
Where n = any number, n(12+3)
How should seven times the sum of a number n and 3 be written as an algebraic expression?
7(n+3)
What is the equation for 3 times the sum of a number n and 7 equals 16?
3 (n + 7) = 16
M times N equals 6300 if m and n are 2 digit multiples of 10 what could m and n be?
m and n are 70 and 90
What is the expresstion of 4 times the difference of a number an 1 is equal to 6 times the sum of a number times 3?
4*(n + 1) = 6*n*3
The 'nth term of an Arithmetic Progression is 3n-2.Find the sum of first n terms.What is the sum of first 10 terms?
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 number 96 can be written as 2m times n where m and n are prime numbers find the value of m and n?
m -5 n-3
Three times the sum of a number and two is negative eight?
3(n+2) = -8 n = -14/3 or 4 and 2/3
