Suppose we have to find the sum of the following AGP series 

\(a + {\rm{ }}\left( {a + d} \right)x\; + {\rm{ }}\left( {a + {\rm{ }}2d} \right){x^2} + {\rm{ }} \ldots .n\) terms  

let us write

\(S = a + {\rm{ }}\left( {a + d} \right)x\; + {\rm{ }}\left( {a + {\rm{ }}2d} \right){x^2} + {\rm{ }} \ldots  \ldots .{T_n}\) ...(1)

Multiplying this equation by \(x\), we get

\(Sx{\rm{ }} = {\rm{ }}ax + {\rm{ }}\left( {a + d} \right){x^2} + {\rm{ }} \ldots ..{T_n}_{-{\rm{ }}1}{x^2} + {T_n}x\) ...(2)

Now subtracting equation (2) from equation (1), we get,

\(S(1 - x) = (a + dx + d{x^2} + d{x^3} + .......n\;\,{\rm{terms)}} - {T_n}x\)

Leaving the first and last terms, we see that remaining \(n-1\) terms are in GP, hence,

\(S(1 - x) = (a - {T_n}x) + \frac{{dx(1 - {x^{n - 1}})}}{{(1 - x)}}\)

\[\bbox[5px, border: 2px solid #0071dc]{S = \frac{{a - [a + (n - 1)d]x}}{{(1 - x)}} + \frac{{dx(1 - {x^{n - 1}})}}{{{{(1 - x)}^2}}}}\]

Example 16: Find the sum of the following series 

\(1{\rm{ }} + {\rm{ }}4x{\rm{ }} + {\rm{ }}7{x^2} + {\rm{ }}10{x^3} + {\rm{ }}13{x^4} + \) ……infinite terms,

where \(0 < x  < 1\)

Solution: Suppose sum of the series is S, then

\(S = {\rm{ }}1{\rm{ }} + {\rm{ }}4x + {\rm{ }}7{x^2} + {\rm{ }}10{x^{3\;}} + {\rm{ }}13{x^4} + {\rm{ }} \ldots \) ..(1)

Multiplying be \(x\), we get,

\(Sx = x + {\rm{ }}4{x^2} + {\rm{ }}7{x^3} + {\rm{ }}10{x^{4\;}} + {\rm{ }}13{x^5} + {\rm{ }}..\) ..(2)

Subtracting equation (2) from equation (1),

\(S\left( {1{\rm{ }}-x} \right){\rm{ }} = {\rm{ }}1{\rm{ }} + {\rm{ }}3x + {\rm{ }}3{x^2} + {\rm{ }}3{x^3} + {\rm{ }}3{x^4} +  \ldots  \ldots . \infty \) terms

\(S( 1 - x) = 1 + \frac{{3x}}{{1 - x}}\) \( \Rightarrow S = \frac{1}{{1 - x}} + \frac{{3x}}{{{{(1 - x)}^2}}}\)

Example 17: find the sum of the series: 

\(1 + \frac{3}{4} + \frac{7}{{16}} + \frac{{15}}{{64}} + \frac{{31}}{{256}} + .....\infty \,\;term\)

Solution: Suppose sum of the given series is \(S\), then

\(S = 1 + \frac{3}{4} + \frac{7}{{16}} + \frac{{15}}{{64}} + \frac{{31}}{{256}} + ......................\infty \;terms\)

\(\frac{3}{4}S = \;\;\;\frac{1}{4} + \frac{3}{{16}} + \frac{7}{{64}} + \frac{{15}}{{256}} + ........................\infty \;terms\)

\(\frac{S}{4} = 1 + \frac{2}{4} + \frac{4}{{16}} + \frac{8}{{64}} + \frac{{16}}{{256}} + ......................\)

\(\frac{S}{4} = 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{{16}} + ............ = \frac{1}{{1 - (1/2)}}\)

\( \Rightarrow  S = 8\).

Example 18:  Find the sum of the series:

\(1 + \frac{{{2^3}}}{5} + \frac{{{3^3}}}{{{5^2}}} + \frac{{{4^3}}}{{{5^3}}} + \frac{{{5^3}}}{{{5^4}}} + \frac{{{6^3}}}{{{5^5}}} + .....\infty \)

Solution: The series is based on difference method

\(S = 1 + \frac{{{2^3}}}{5} + \frac{{{3^3}}}{{{5^2}}} + \frac{{{4^3}}}{{{5^3}}} + \frac{{{5^3}}}{{{5^4}}} + \frac{{{6^3}}}{{{5^5}}} + .......\) (1)

\(\frac{S}{5} = \frac{1}{5} + \frac{{{2^3}}}{{{5^2}}} + \frac{{{3^3}}}{{{5^3}}} + \frac{{{4^3}}}{{{5^4}}} + \frac{{{5^3}}}{{{5^5}}} + ..........\).. (2)

Subtract (2) from (1) we get

\(\frac{4}{5}S = 1 + \frac{7}{5} + \frac{{19}}{{{5^2}}} + \frac{{37}}{{{5^3}}} + \frac{{61}}{{{5^4}}} + \frac{{91}}{{{5^5}}} + ...\) (3)

Multiplying equation (3) by (1/5) and subtracting from equation (3),

\(\frac{{16}}{{25}}S = 1 + \frac{6}{5} + \frac{{12}}{{{5^2}}} + \frac{{18}}{{{5^3}}} + .......\) (4)

Multiplying equation (4) by \(1/5\) and subtracting from equation (4),

\(\frac{{64}}{{125}}S = 1 + \frac{5}{5} + \frac{6}{{25}} + \frac{6}{{125}} + ......\)

\(\frac{{64}}{{125}}S = 1 + 1 + \frac{{6/25}}{{1 - 1/5}} = \frac{{23}}{{10}}\)

\( \Rightarrow S = \frac{{125}}{{64}} \times \frac{{23}}{{10}} = \frac{{575}}{{128}}\)

Example 19: Find the sum of the series:

\(2 + 12 + 36 + 80 + 150 + 252 + 392 +  … 30\) terms. 

Solution: We know that if the difference of the consecutive terms of a series is constant then the series is in AP and it’s nth term is linear in terms of \(n\). Similarly if the difference of difference (2nd difference) is constant, then the \(nth\) term is quadratic in terms of \(n\). In this case third difference is constant hence the \(nth\) term is a cubic expression in terms on n.

Suppose \({T_n} = a{n^3} + b{n^2} + cn + d\)

\({T_1} = a + b + c + d = 2\) (1)

\({T_2} = 8a + 4b + 2c + d = 12\) (2)

\({T_2} = 27a + 9b + 3c + d = 36\) (3)

\({T_4} = 64a + 16b + 4c + d = 80\) (4)

Writing \({T_2}-{T_1},{T_3}-{T_2}\) and \({T_4}-{T_3}\)

\(7a + 3b + c = 10\)

\(19a + 5b + c = 24\)

\(37a + 7b + c = 44\)

Solving these equation we get,

\(a =1, b = 1, c = 0, d = 0\) 

\({T_n} = {n^3} + {n^2}\)

\(\sum\limits_{n = 1}^{n = 30} {{T_n}}  = {\left( {\frac{{n(n + 1)}}{2}} \right)^2} + \left( {\frac{{n(n + 1)(2n + 1)}}{6}} \right)\) = 225680.

(Note that this method is another form of difference method. Before applying this method, students should inspect the series carefully, some times the pattern in the series is very explicit and does not require all the detailed analysis discussed above. In the given series, 

\(2{\rm{ }} = {\rm{ }}{1^3} + {\rm{ }}{1^2},{\rm{ }}12{\rm{ }} = {\rm{ }}{2^3} + {\rm{ }}{2^2},{\rm{ }}36{\rm{ }} = {\rm{ }}{3^3} + {\rm{ }}{3^2}\) and so on, hence the \(nth\) term of the series is \({n^3} + {\rm{ }}{n^2}\).)