Question

Find the sum to n terms in the following series.
2.3.4 + 3.4.5 + .........

Please explain me in detail, thanQ!!

Answer 1

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?

Answer 2

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!

Answer 3

Wow, I thought for sure going to the 3rd difference would have been enough.

"Nasty" question!, but nice one.

Answer 4

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