Hence the sum =A+ 2n 3 +n^2 + 1 3 n^3.
To find A, put n = 1; the series then reduces to its first term, and 2 = A + 2 or A = 0
Hence 12+23+34+ +n(n+1)=1 3n(n+1)(n +2).
Note. It will be seen from this example that when the nth term is a rational integral function of n, it is sufficient to assume for the sum a function of x which is of one dimension higher than the nth term of the series.
Ex. 2. Find the conditions that x^3 + px^2 + qx + r may be divisible by x^2+ ax +b.
The quotient will contain two terms; namely, x and a term independent of x. Hence, we assume
x^3 + px^2 + qx +r =(x + k) (x^2 + ax + b).
Equating the coefficients of the like powers of x, we have
k+a = p, ak+b = q, kb=r.
From the last equation k = r b hence by substitution we obtain
r b + a =p, and ar b+b=q;
that is, r=b(p-a), and ar=b(q-b);
which are the conditions required.
EXAMPLES XLII. a.
Find by the method of Undetermined Coefficients the sum of
1. 1^2 + 3^2 + 5^2 + 7^2 +.. to n terms. 2. 1-2-3+2.3.4+3.4-5+.. to n terms. 3. 1 2^2 +2 3^2+3 4^2 + 4 5^2 +... to n terms. 4. 1^3 + 3^3 + 5^3 + 7^3 + to n terms. 5. 1^4 + 2^4 + 3^4+ 4^4 + to n terms.
6. Find the condition that x^3 - 3px +2q may be divisible by a factor of the form x^2 + 2ax + a^2.
7. Find the conditions that ax^3 + bx^2 + cx +d may be a perfect cube.
