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There is no solution to the question as asked.
If the sum of n terms is 2n+1 then the sum of n+1 terms, using the same formula, is 2*(n+1)+1 = 2n+2+1 = 2n+3
So the (n+1)th term is sum to n+1 minus sum to n = (2n+3) - (2n+1) = 2
So each term is 2. But if each term is 2, then the sum of n terms must be even. The sum is clearly odd - which leads to a contradiction.
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What is the sum of the 3rd and 5th term in the sequence where the nth term is 3n-4?
16
What is the sum of the positive even integers less than one hundred?
This can be found out by using the formula for sum of n terms of an arithmetic progression. here, n is not known. a=2 d=2 (since it's even) nth term=98 (last even number less than 100) using formula for nth term, nth term=a+(n-1)d 98=2+(n-1)2 therefore, n=49 so, sum of n terms=n/2[2*a+(n-1)d] putting n=49, a=2, and d=2, sum=2450
The sum of the first 5 terms of an arithmetic series is 85. The sum of the first 6 terms is 123. Determine the first four terms of the series.?
Suppose the first term is a, the second is a+r and the nth is a+(n-1)r. Then the sum of the first five = 5a + 10r = 85 and the sum of the first six = 6a + 15r = 123 Solving these simultaneous equations, a = 3 and r = 7 So the first four terms are: 3, 10, 17 and 24
Find the sum of the first 48 terms of an aritmetic sequance 2 4 6 8?
To find the sum of the first 48 terms of an arithmetic sequence, we can use the formula for the sum of an arithmetic series: Sn = n/2 * (a1 + an), where Sn is the sum of the first n terms, a1 is the first term, and an is the nth term. In this case, a1 = 2, n = 48, and an = 2 + (48-1)*2 = 96. Plugging these values into the formula, we get: S48 = 48/2 * (2 + 96) = 24 * 98 = 2352. Therefore, the sum of the first 48 terms of the given arithmetic sequence is 2352.
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
Give the simple formula for the nth term of the following arithmetic sequence What if your answer will be of the form an plus b?
If the formula for additional terms was the summation of the term before it, the nth term of the series would be the sum of all terms prior. In other words it would be the summation of a through n minus 1.
What is the sum of first six terms of a sequence whose nth term is 8 - n?
nth term is 8 - n. an = 8 - n, so the sequence is {7, 6, 5, 4, 3, 2,...} (this is a decreasing sequence since the successor term is smaller than the nth term). So, the sum of first six terms of the sequence is 27.
How can you find the nth partial?
The Nth partial sum is the sum of the first n terms in an infinite series.
What is the rule for the sum of the first n terms of a geometric series?
Suppose the nth term is = arn where n = 1,2,3, ... Then the sum to the nth term is a*(rn+1 - 1)/(r - 1) or, equivalently, a*(1 - rn+1)/(1 - r)
What is the sum of the 3rd and 5th term in the sequence where the nth term is 3n-4?
16
What is the sum of the positive even integers less than one hundred?
This can be found out by using the formula for sum of n terms of an arithmetic progression. here, n is not known. a=2 d=2 (since it's even) nth term=98 (last even number less than 100) using formula for nth term, nth term=a+(n-1)d 98=2+(n-1)2 therefore, n=49 so, sum of n terms=n/2[2*a+(n-1)d] putting n=49, a=2, and d=2, sum=2450
Is 9 plus 3u plus 5u plus 4v a product of four factors?
No. It is a sum of four terms which can be simplified to a sum of three terms.
What is the nth term for 0 1 1 2 3 5 8 13?
This is the Fibonacci sequence, where the number is the sum of the two preceding numbers. The nth term is the (n-1)th term added to (n-2)th term
How do you find the C programming of the sum of the series 5 plus 55 plus 555 plus . plus n terms?
Find the Sum to n terms of the series 5 5+55+555+ +n Terms
The sum of the first 5 terms of an arithmetic series is 85. The sum of the first 6 terms is 123. Determine the first four terms of the series.?
Suppose the first term is a, the second is a+r and the nth is a+(n-1)r. Then the sum of the first five = 5a + 10r = 85 and the sum of the first six = 6a + 15r = 123 Solving these simultaneous equations, a = 3 and r = 7 So the first four terms are: 3, 10, 17 and 24
Find the sum of the first 48 terms of an aritmetic sequance 2 4 6 8?
To find the sum of the first 48 terms of an arithmetic sequence, we can use the formula for the sum of an arithmetic series: Sn = n/2 * (a1 + an), where Sn is the sum of the first n terms, a1 is the first term, and an is the nth term. In this case, a1 = 2, n = 48, and an = 2 + (48-1)*2 = 96. Plugging these values into the formula, we get: S48 = 48/2 * (2 + 96) = 24 * 98 = 2352. Therefore, the sum of the first 48 terms of the given arithmetic sequence is 2352.
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
