Question

Find the sum of n terms of the sequence
$(x + y),({x^2} + xy + {y^2}),({x^3} + {x^2}y + x{y^2} + {y^3}),....$ to n terms.

Answer 1

Here we will multiply and divide each of the terms by $\left( {x - y} \right)$ and then use the formula for sum of terms in GP to get the desired answer.
The sum of n terms of a GP is given by:-
$S = \dfrac{{a\left( {{r^n} - 1} \right)}}{{r - 1}}$ where a is the first term and r is the common ratio.

Complete step-by-step answer:
The given terms are:-
$(x + y),({x^2} + xy + {y^2}),({x^3} + {x^2}y + x{y^2} + {y^3}),....$
Now adding these terms we get:-
\Rightarrow$$$(x + y) + ({x^2} + xy + {y^2}) + ({x^3} + {x^2}y + x{y^2} + {y^3}) + ....$$ Now multiplying and dividing each term by $$\left( {x - y} \right)$$ we get:- \Rightarrow\dfrac{1}{{x - y}}\left[ {(x + y)\left( {x - y} \right) + \left( {x - y} \right)({x^2} + xy + {y^2}) + \left( {x - y}\right)({x^3} + {x^2}y + x{y^2} + {y^3}) + ....} \right]$$ Now we know that, $\Rightarrow{x^2} - {y^2} = (x + y)\left( {x - y} \right) Also, $\Rightarrow$$${x^3} - {y^3} = \left( {x - y} \right)({x^2} + xy + {y^2})
Hence, substituting the values we get:-
\Rightarrow$$$S = \dfrac{1}{{\left( {x - y} \right)}}\left[ {{x^2} - {y^2} + {x^3} - {y^3} + {x^4} - {y^4} + ....} \right]$$ Splitting the terms we get:- \RightarrowS = \dfrac{1}{{\left( {x - y} \right)}}\left[ {{x^2} + {x^3} + {x^4} + ....{\text{n terms}}} \right] - \dfrac{1}{{\left( {x - y} \right)}}\left[ {{y^2} + {y^3} + {y^4} + ....{\text{n terms}}} \right]$$ Now we can see that both x and y are forming a GP Hence we will apply the formula of n terms of a GP which is given by:- $$S = \dfrac{{a\left( {{r^n} - 1} \right)}}{{r - 1}}$$ where a is the first term and r is the common ratio. Hence, applying the formula we get:- $\RightarrowS = \dfrac{1}{{\left( {x - y} \right)}}\left[ {\dfrac{{{x^2}\left( {{x^n} - 1} \right)}}{{x - 1}}} \right] -\dfrac{1}{{\left( {x - y} \right)}}\left[ {\dfrac{{{y^2}\left( {{y^n} - 1} \right)}}{{y - 1}}} \right] Simplifying it we get:- $\Rightarrow$$$S = \dfrac{1}{{\left( {x - y} \right)}}\left[ {\dfrac{{{x^2}\left( {{x^n} - 1} \right)}}{{x - 1}} -\dfrac{{{y^2}\left( {{y^n} - 1} \right)}}{{y - 1}}} \right]
Hence the sum of the given sequence is:-

$S = \dfrac{1}{{\left( {x - y} \right)}}\left[ {\dfrac{{{x^2}\left( {{x^n} - 1} \right)}}{{x - 1}} - \dfrac{{{y^2}\left( {{y^n} - 1} \right)}}{{y - 1}}} \right]$

Note: Students should note that in geometric progression two consecutive have the same common ratio as that other two consecutive terms of a series.
Also, students should note that the sum of n terms of a series in GP is given by:-
$S = \dfrac{{a\left( {{r^n} - 1} \right)}}{{r - 1}}$ when $r > 1$
Where a is the first term and r is the common ratio.
And,
$S = \dfrac{{a\left( {1 - {r^n}} \right)}}{{1 - r}}$ when $r < 1$
Where a is the first term and r is the common ratio.