Here you will learn what is agp (Arithmetico Geometric Series) and how to solve agp series.
Let’s begin –
What is AGP (Arithmetico – Geometric series)
A series, each term of which is formed by multiplying the corresponding term of an A.P. & G.P. is called the Arithmetico-Geometric Series, e.g. 1 + 3x + 5\(x^2\) + 7\(x^3\) + ……..
Here 1, 3, 5, ……. are in A.P. & 1, x, \(x^2\), \(x^3\) …….. are in G.P.
(a) Sum of N terms of an arithmetico-geometric series :
Let \(S_n\) = a + (a + d)r + …….. + [a + (n-1)d]\(r^{n-1}\)
then \(S_n\) = \(a\over {1-r}\) + \(dr({1}-{r}^{n-1})\over (1-r)^2\) – \([a + (n-1)d]r^2\over {1-r}\), r \(\ne\) 1
(b) Sum of Infinity :
If 0 < |r| < 1 & n \(\rightarrow\) \(\infty\) \(\displaystyle \lim_{x \to \infty}\) \(r^n\) = 0, \(S_{\infty}\) = \(a\over {1-r}\) + \(dr\over (1-r)^2\)
Example : Find the sum of series 4 – 9x + 16\(x^2\) – 25\(x^3\) + 36\(x^4\) – 49\(x^5\) + ……. \(\infty\)
Solution : Let S = 4 – 9x + 16\(x^2\) – 25\(x^3\) + 36\(x^4\) – 49\(x^5\) + ……. \(\infty\)
-Sx = -4x + 9\(x^2\) – 16\(x^3\) + 25\(x^4\) – 36\(x^5\) + ……. \(\infty\)
On Subtraction, we get
S(1 + x) = 4 – 5x + 7\(x^2\) – 9\(x^3\) + 11\(x^4\) – 13\(x^5\) + ……. \(\infty\)
-S(1 + x)x = -4x + 5\(x^2\) – 7\(x^3\) + 9\(x^4\) – 11\(x^5\) + ……. \(\infty\)
On Subtraction, we get
S\((1 + x)^2\) = 4 – x + 2\(x^2\) – 2\(x^3\) + 2\(x^4\) – 2\(x^5\) + ……. \(\infty\)
= 4 – x + 2\(x^2\)(1 – x + \(x^2\) – \(x^3\) + ……. \(\infty\)) = 4 – x + \(2x^2\over {1+x}\)
= \({4 + 3x + x^2}\over {1+x}\)
S = \({4 + 3x + x^2}\over (1+x)^3\)
RESULTS
(a) \({\sum}_{r=1}^{n}\)r = \(n(n + 1)\over 2\) (sum of the first n natural numbers)
(b) \({\sum}_{r=1}^{n} r^2\) = \(n(n + 1)(2n + 1)\over 6\) (sum of the squares of the first n natural numbers)
(c) \({\sum}_{r=1}^{n} r^3\) = \(n^2(n + 1)^2\over 4\) (sum of the cubes of the first n natural numbers)
(d) \({\sum}_{r=1}^{n} r^4\) = \(n(n + 1)(2n + 1)(3n^2 + 3n -1)\over 30\)
(e) \({\sum}_{r=1}^{n}\)(2r – 1) = \(n^2\) (sum of the first n odd natural numbers)
(f) \({\sum}_{r=1}^{n}\)2r = n(n + 1) (sum of the first n even natural numbers)
Note :
If \(n^{th}\) terms of a sequence is given by \(T_n\) = \(an^3\) + \(bn^2\) + cn + d where a, b, c, d are constants,
then sum of n terms \(S_n\) = \(\sum T_n\) = a\(\sum n^3\) + b\(\sum n^2\) + c\(\sum n\) + \(\sum d\)
This can be evaluated using the above results.
Example : Sum upto 16 terms of the series \(1^3\over 1\) + \(1^3 + 2^3\over 1 + 3\) + \(1^3 + 2^3 + 3^3\over 1 + 3 + 5\) +…….. is
Solution : \(t_n\) = \(1^3 + 2^3 + 3^3 + …. + n^3\over 1 + 3 + 5 + ….. + (2n – 1)\)
= \({n^2(n + 1)^2\over 4}\over n/2{[2+2(n – 1)]}\) = \({n^2(n + 1)^2\over 4}\over n^2\)
= \({(n + 1)^2\over 4}\) = \(n^2\over 4\) + \(n\over 4\) + \(1\over 4\)
\(\therefore\) \(S_n\) = \(\sum t_n\) + \(1\over 4\)\(\sum n^2\) + \(1\over 2\)\(\sum n\) + \(1\over 4\)\(\sum 1\)
= \(1\over 4\).\(n(n + 1)(2n + 1)\over 6\) + \(1\over 2\).\(n(n + 1)\over 2\) + \(1\over 4\).n
\(\therefore\) \(S_{16}\) = \(16.17.33\over 24\) + \(16.17\over 4\) + \(16\over 4\) = 446.