Find the sum to n terms in the following series.
2.3.4 + 3.4.5 + .........
Please explain me in detail, thanQ!!
I will assume the . is a multiplication
sum(1) = 24
Sum(2) = 24 + 60 = 84
sum(3) = 84 + (4)(5)(6) = 204
sum(4) = 204 + (5)(6)(7) = 414
sum(5) = 414 + (6)(7)(8) = 750
....
investigate the sequence: 24 84 204 414 750 ..
I took differences, and the 1st, 2nd, and 3rd differences are not constant.
So the terms cannot be expressed either as linear, quadratic or cubic expressions.
At this point I see no pattern.
Is this a question from the topic of induction?
It turns out that it is a quartic!
The fourth difference is 6.
The sum to nth term of the given series is therefore
(x*(x+5)*(x^2+5*x+10))/4 which is a little awkward.
Had the series started at
1.2.3 + 2.3.4 + 3.4.5 +..., then
S(n)=n(n+1)(n+2)(n+3)/4
a nice and easy to remember expression!
Wow, I thought for sure going to the 3rd difference would have been enough.
"Nasty" question!, but nice one.
Naturally the next question is, how do we get the coefficients of the resulting polynomial?
We first calculate the numerical values of g(x) and the differences until it is constant. We will take advantage of the calculations Reiny has already done:
sum(6) = 750 + (7)(8)(9) = 1254
sum(5) = 414 + (6)(7)(8) = 750
sum(4) = 204 + (5)(6)(7) = 414
sum(3) = 84 + (4)(5)(6) = 204
Sum(2) = 24 + 60 = 84
sum(1) = 24
The differences are then:
x sum(x) Δ1 Δ2 Δ3 Δ4
6 1254
5 750 504
4 414 336 168
3 204 210 126 42
2 84 120 90 36 6
1 24 60 60 30 6
It would be advisable to calculate one or two more terms to ensure that the polynomial is indeed of degree 4 (i.e. Δ5 = constant).
With the degree of the polynomial established as 4, without loss of generality, we can assume the polynomial to be:
g(x)=a4x^4+a3x^3+a2x^2+a1x+a0
where a4,a3,a2,a1,a0 are constants to be determined.
Δ4(1) was calculated by:
g(5)-4g(4)+6g(3)-4g(2)+g(1)=6
expanding g(i) for i=1 to 5, and after simplification, we get:
24a4 = 6, or
a4=(1/4)
Similarly, we expand
g(4)-3g(3)+3g(2)-g(1) = 30
which gives after simplification:
60a4+6a3 = 30
from which we can solve for a3:
a3=(30-(60/4))/6= (5/2)
Proceeding similarly,
g(3)-2g(2)+g(1) = 60 =>
50a4+12a3+2a2 = 60
a2 = (60-50/4-12(5/2))/2 = (35/4)
and
g(2)-g(1) = 60 =>
15a4+7a3+3a2+a1 = 60
a1 = 60 - 15/4 - 7(5/2) - 3(35/4) = 25/2
finally,
g(1)=a4+a3+a2+a1+a0=24
a0=24-(1/4)-(5/2)-(35/4)-25/2 = 0
Therefore:
g(x)=(1/4)x^4+(5/2)x^3+(35/4)x^2+(25/2)x
or
g(x)=x(x^3+10x^2+35x+50)/4
which after factorization
=x(x+5)(x^2+5x+10)/4
To find the sum of the series, we need to understand the pattern and come up with a general formula.
Let's look at the given series: 2.3.4 + 3.4.5 + ...
Here, each term consists of three consecutive numbers being multiplied together. The first term is 2*3*4, the second term is 3*4*5, and so on.
To find a general formula for the nth term, we can observe that:
- The first number of each term increases by 1 (starting from 2, then 3, 4, and so on).
- The second number of each term is the first number plus 1.
- The third number of each term is the second number plus 1.
From this observation, we can say that the nth term can be expressed as:
(n) * (n+1) * (n+2)
To find the sum of the first n terms, we need to add all these terms together. We can use the summation symbol Σ to represent the sum:
Sum = Σ (n) * (n+1) * (n+2)
Now, we can expand this sum expression and simplify it.
Step 1: Expand the expression
Sum = 2*3*4 + 3*4*5 + 4*5*6 + ...
Step 2: Group the common factors
Sum = 2*(3*4) + 3*(4*5) + 4*(5*6) + ...
Step 3: Apply the distributive property
Sum = 2*3*(3+1) + 3*4*(4+1) + 4*5*(5+1) + ...
Step 4: Simplify
Sum = 6*(3+1) + 12*(4+1) + 20*(5+1) + ...
Step 5: Distribute and collect like terms
Sum = 6*3 + 6*1 + 12*4 + 12*1 + 20*5 + 20*1 + ...
Step 6: Simplify further
Sum = 18 + 6 + 48 + 12 + 100 + 20 + ...
Now, we have simplified the sum expression. We can see that the terms are in an arithmetic sequence with a common difference of 6.
To find the sum of the series, we can use the formula for the sum of an arithmetic series:
Sum = (n/2) * (first term + last term)
In this case, the first term is 18, and the common difference is 6. So, the last term will be 18 + (n-1)*6.
Therefore, the sum of the series is:
Sum = (n/2) * (18 + (18 + (n-1)*6))
Simplifying further:
Sum = (n/2) * (36 + 6n - 6)
Sum = (n/2) * (30 + 6n)
Now, we have the general formula for finding the sum to n terms in the given series: Sum = (n/2) * (30 + 6n).
To find the sum, substitute the value of n and simplify the expression.