If a=∑n=0∞ xn,b=∑n=0∞ yn,c=∑n=0∞ (xy)n,where |x|,|y|<1,then

a=∑n=0∞ xn .....(i)b=∑n=0∞ yn ...(ii) c=∑n=0∞ (xy)n ....(iii) From Eq. (i),a=∑n=0∞ xn=1+x+x2+x3+⋯+∞=11−x [∵|x|<1]… (iv) b=∑n=0∞ yn=1+y+y2+y3+⋯+∞=11−y [∵|y|<1]… (v) From Eq. (iv), a=11−x⇒ a(1−x)=1⇒ a−ax=1⇒ x=a−1a .......(iv)From Eq. (v),b=11−y⇒ y=b−1b ......(vii)Now, c=∑n=0∞ (xy)nc=1+xy+(xy)2+(xy)3+⋯+∞=11−xy [∵|x|<1,|y|<1⇒|xy|<1]=11−a−1ab−1bFrom Eqs. (vi) and (vii) c=1ab−(a−1)(b−1)ab⇒c=aba⋅b−{ab−a−b+1}⇒c=abab−ab+a+b−1⇒c=aba+b−1⇒ca+cb−c=ab⇒ac+bc=ab+c

If a=∑n=0∞ xn,b=∑n=0∞ yn,c=∑n=0∞ (xy)n,where |x|,|y|<1,then

a=∑n=0∞ xn .....(i)b=∑n=0∞ yn ...(ii) c=∑n=0∞ (xy)n ....(iii) From Eq. (i),a=∑n=0∞ xn=1+x+x2+x3+⋯+∞=11−x [∵|x|<1]… (iv) b=∑n=0∞ yn=1+y+y2+y3+⋯+∞=11−y [∵|y|<1]… (v) From Eq. (iv), a=11−x⇒ a(1−x)=1⇒ a−ax=1⇒ x=a−1a .......(iv)From Eq. (v),b=11−y⇒ y=b−1b ......(vii)Now, c=∑n=0∞ (xy)nc=1+xy+(xy)2+(xy)3+⋯+∞=11−xy [∵|x|<1,|y|<1⇒|xy|<1]=11−a−1ab−1bFrom Eqs. (vi) and (vii) c=1ab−(a−1)(b−1)ab⇒c=aba⋅b−{ab−a−b+1}⇒c=abab−ab+a+b−1⇒c=aba+b−1⇒ca+cb−c=ab⇒ac+bc=ab+c

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

If $a = \sum_{n = 0}^{\infty} x^{n}, b = \sum_{n = 0}^{\infty} y^{n}, c = \sum_{n = 0}^{\infty} (xy)^{n},$ where $| x |, | y | < 1$ ,then

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$abc = a + b + c$

$ab + bc = ac + b$

$ac + bc = ab + c$

$ab + ac = bc + a$

answer is C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\begin{array}{l} a = \sum_{n = 0}^{\infty} x^{n} . . . . . (i) \\ b = \sum_{n = 0}^{\infty} y^{n} . . . (ii) \\ c = \sum_{n = 0}^{\infty} (xy)^{n} . . . . (iii) \end{array}$

From Eq. (i),

$\begin{aligned} a = \sum_{n = 0}^{\infty} x^{n} = 1 + x + x^{2} + x^{3} + \dots + \infty \\ = \frac{1}{1 - x} [∵ | x | < 1] \dots (iv) \\ b = \sum_{n = 0}^{\infty} y^{n} = 1 + y + y^{2} + y^{3} + \dots + \infty \\ = \frac{1}{1 - y} [∵ | y | < 1] \dots (v) \end{aligned}$

From Eq. (iv), $a = \frac{1}{1 - x}$

$\begin{array}{lrl} \Rightarrow a (1 - x) = 1 \Rightarrow a - ax = 1 \\ \Rightarrow x = \frac{a - 1}{a} . . . . . . . (iv) \end{array}$

From Eq. (v), $b = \frac{1}{1 - y}$

$\Rightarrow y = \frac{b - 1}{b} . . . . . . (vii)$

Now, $c = \sum_{n = 0}^{\infty} (xy)^{n}$

$\begin{array}{l} c = 1 + xy + (xy)^{2} + (xy)^{3} + \dots + \infty \\ = \frac{1}{1 - xy} [∵ | x | < 1, | y | < 1 \Rightarrow | xy | < 1] \end{array}$

$= \frac{1}{1 - (\frac{a - 1}{a}) (\frac{b - 1}{b})}$

From Eqs. (vi) and (vii)

$c = \frac{1}{\frac{ab - (a - 1) (b - 1)}{ab}}$

$\begin{array}{l} \Rightarrow c = \frac{ab}{a \cdot b - {ab - a - b + 1}} \\ \Rightarrow c = \frac{ab}{ab - ab + a + b - 1} \\ \Rightarrow c = \frac{ab}{a + b - 1} \\ \Rightarrow ca + cb - c = ab \Rightarrow ac + bc = ab + c \end{array}$