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 will do the first one
We were lucky to have the general term, in most cases you have to find that first
Secondly you have to know some basic summation formulas
∑ k ,where k = 1 to n is n(n+1)/2
∑ k^2 , where k = 1 to n is n(n+1)(2n+1)/6
∑ c , where k = 1 to n is kc , where c was a constant
So we want ∑ (2n-1)^2
= ∑(4n^2 - 4n + 1)
= 4∑ n^2 - 4∑ n + ∑ 1
= 4n(n+1)(2n+1)/6 - 4n(n+1)/2 + n(1)
= (4n(n+1)(2n+1) - 12n(n+1) + 6n)/6
= ..
I will let you simplify this
Finding the sum of the first 2 terms of the series with the above formula won't get you the right answer
yes it will
Sum(1) = (4(2)(3) - 12(2) + 6)/6 = 1
sum(2) = (8(3)(5) - 24(3) + 12)/6 = 10
sum(3) = 12(4)(7) - 36(4) + 18)/6 = 35
etc.
It works! I am sure!
1) To find the sum of the n terms of the series Sn = 1^2 + 3^2 + 5^2 + ... + (2n-1)^2, we can use the formula for the sum of the squares of the first n natural numbers.
First, let's observe the series and see if we can find a pattern:
1^2 = 1
3^2 = 9
5^2 = 25
...
We can see that each term in the series is the square of an odd number. The first term is 1^2 = 1, the second term is 3^2 = 9, the third term is 5^2 = 25, and so on. We can write the nth term as (2n-1)^2.
The sum of the squares of the first n natural numbers is given by the formula:
Sn = n(n+1)(2n+1)/6
In this case, we have (2n-1) in place of n. So, the sum of the n terms can be calculated as:
Sn = (2n-1)(2n-1+1)(2(2n-1)+1)/6
= (2n-1)(2n)(4n-1)/6
= (4n^3 - 6n^2 + 2n)/6
= (2n(2n^2 - 3n + 1))/6
= n(2n^2 - 3n + 1)/3
Therefore, the sum of the n terms of the series Sn = 1^2 + 3^2 + 5^2 + ... + (2n-1)^2 is given by Sn = n(2n^2 - 3n + 1)/3.
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 can also identify a pattern and use it to find a formula for the nth term.
Let's observe the series and find the pattern:
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 see a repeating pattern in the denominators: 6, 4, 6, 8. The numerators also follow a pattern: 1, 3, 5, 7. The nth term can be written as (2n-1)/[(n+1)(n+2)(n+3)].
To find the sum to n terms, S_n, we will express each term in fractional form and then add them up.
S_n = (2n-1)/[(1+1)(1+2)(1+3)] + (2n+1)/[(2+1)(2+2)(2+3)] + (2n+3)/[(3+1)(3+2)(3+3)] + ...
Now, let's simplify the expression:
S_n = (2n-1)/6 + (2n+1)/24 + (2n+3)/60 + ...
We can observe that each term in the series follows a pattern. The numerators increase by 2 with each subsequent term, and the denominators increase by 4 and 2 alternately. Let's express this pattern using a general term formula.
The general term (T) for this series is:
T = [(2n + (2k-1)) / (k(k+1)(k+2))]
Now, we can proceed to summing up this series.
3) To find the sum to n terms of the series 1.3.5 + 2.4.6 + 3.5.7 + ..., we can also identify a pattern and use it to find a formula for the nth term.
Observing the series, we can see that the first term is the product of the numbers 1, 3, and 5. The second term is the product of the numbers 2, 4, and 6. The third term is the product of the numbers 3, 5, and 7. From this pattern, we can deduce that the nth term is the product of the numbers (n, n+2, n+4).
To find the sum to n terms, S_n, we will express each term in the series and then add them up.
S_n = (1 x 3 x 5) + (2 x 4 x 6) + (3 x 5 x 7) + ...
We can observe that each term in the series follows a pattern. The first number starts with n and increases by 1 with each subsequent term. The second and third numbers start with n+2 and n+4, respectively, and also increase by 2 with each subsequent term.
Let's express this pattern using a general term formula.
The general term (T) for this series is:
T = (n x (n+2) x (n+4))
Now, we can proceed to summing up this series by adding up the general terms.