1).Find the sum of the n terms of the series Sn=1^2 + 3^2 +5^2+...+(2n-1)^2
2).Find the sum to n terms of 1/(1.2.3) + 3/(2.3.4) + 5/(3.4.5) + 7/(4.5.6) +....
3) Sum to n terms,the series 1.3.5 + 2.4.6 + 3.5.7 +.
I answered the first part in your previous post
Still messing around with #2
#3,
∑ k(k+2)(k+4) , where k = 1 to n
if you expand this you will get a cubic , so you will have to have
∑ k^3 for k=1 to n equal to (1/4)n^2(n+1)^2
( I had to look that one up)
for the square and first degree terms use the ones I gave you in #1
good luck
n
∑ (2k-1)/[k(k+1)(k+2) =
k=1
n(3n+1)
--------------
4(n+1)(n+2)
1) To find the sum of the n terms of the series Sn=1^2 + 3^2 + 5^2 + ... + (2n-1)^2, we need to first recognize that this is an arithmetic series with a common difference of 2.
Let's break down the series into individual parts:
1^2 = 1
3^2 = 9
5^2 = 25
...
We can observe that each term in the series is obtained by squaring the odd numbers, starting from 1 and increasing by 2 each time.
Now, to find the sum of the n terms, we can use the formula for the sum of an arithmetic series:
Sn = (n/2) * (2a + (n-1)d)
In this case, the first term (a) is 1, and the common difference (d) is 2. Substituting these values into the formula:
Sn = (n/2) * (2(1) + (n-1)(2))
= (n/2) * (2 + 2n - 2)
= (n/2) * (2n)
= n^2
Therefore, the sum of the n terms of the given series is n^2.
2) To find the sum to n terms of the series 1/(1.2.3) + 3/(2.3.4) + 5/(3.4.5) + 7/(4.5.6) + ..., we need to recognize the pattern and express it in a simplified form.
Let's break down the individual terms:
1/(1.2.3) = 1/6
3/(2.3.4) = 1/4
5/(3.4.5) = 1/6
7/(4.5.6) = 1/8
...
We can observe that each term in the series has a denominator formed by the product of three consecutive numbers, and the numerator is obtained by adding 2 to the denominator.
Now, to find the sum to n terms of this series, we can generate a general term for the numerator and denominator:
Numerator: 2n + 1
Denominator: (n+1)(n+2)(n+3)
Therefore, the n-th term of the series can be expressed as (2n + 1) / (n+1)(n+2)(n+3).
Now, to find the sum to n terms, we can use partial fractions or telescoping series technique. However, given the complexity of the series, it is unlikely to find a closed-form expression for the sum. Instead, we can use numerical methods or calculation tools to evaluate the sum for a given value of n.
3) To find the sum to n terms of the series 1.3.5 + 2.4.6 + 3.5.7 + ..., we need to recognize the pattern and express it in a simplified form.
Let's break down the individual terms:
1.3.5 = 15
2.4.6 = 48
3.5.7 = 105
...
We can observe that each term in the series is obtained by multiplying consecutive odd and even numbers, such as (1 * 3 * 5), (2 * 4 * 6), (3 * 5 * 7), and so on.
Now, to find the sum to n terms of this series, we can generate a general term:
(2n-1)(2n)(2n+1)
Therefore, the n-th term of the series can be expressed as (2n - 1)(2n)(2n + 1).
Now, to find the sum to n terms, we can either use algebraic manipulation techniques or use summation formulas based on calculus. However, since the series is quite complex, it might not be easy to find a closed-form expression for the sum. Instead, numerical methods or calculation tools can be used to find the sum for a given value of n.