Bonaparte said:Its a lot of messing around with, but I'll give you the basic idea, note that when you expand, the integers turn out to be sum of squares, that is (n-1)(n)(2n-1)/6. The rn turn out to be (n-1)rn. You just do this to the different terms to get the final thing, which then you might be able to factor out. Try doing this and post your result, shouldn't be too hard.
Thanks, Bonaparte
Vineeth T said:Can you explain it more clearly?
Also the source of this problem is from a book called "HIGHER ALGEBRA" by Hall&Knight.
If you have this book see the answer (only the final result is given) in pg:328.Q no:27
The answer even has factorial notations in it.
A sequence is a list of numbers arranged in a specific order. Each number in the sequence is called a term and the position of the term in the sequence is called its index.
A series is the sum of all the terms in a sequence. It is denoted by the Greek letter sigma (∑) and the starting and ending index of the terms are usually indicated below and above the sigma symbol respectively.
An arithmetic sequence is a sequence where the difference between any two consecutive terms is constant. A geometric sequence is a sequence where the ratio between any two consecutive terms is constant. In other words, in an arithmetic sequence, the terms increase or decrease by a fixed amount while in a geometric sequence, the terms increase or decrease by a fixed ratio.
The formula for finding the nth term of an arithmetic sequence is:
an = a₁ + (n-1)d
where an is the nth term, a₁ is the first term, and d is the common difference.
The formula for finding the sum of a finite arithmetic series is:
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.
The formula for finding the sum of a finite geometric series is:
Sn = (a₁(1-rⁿ))/(1-r)
where Sn is the sum of the first n terms, a₁ is the first term, and r is the common ratio.