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If 1+a+a2+………∞=x and 1+b+b2+……∞=y then 1+ab+a2b2+………∞ = x is given by ________.
- a)(xy)/(x+y-1)
- b)(xy)/(x-y-1)
- c)(xy)/(x+y+1)
- d)None
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
If 1+a+a2+………∞=x and 1+b+b2+…&hellip...
ANSWER :- a
Solution :- Given, x=1+a+a^2+......∞
Since this is a infinite G.P. series, where, (first term)=1 and (common difference)=a,
So, x = 1/(1−a)
⇒ x−ax=1
⇒ ax=x−1
⇒ a=(x−1)/x
Similarly, y=1+b+b^2 +......∞ is a infinite G.P. series, where, (first term)=1 and
(common difference)=b,
So, y = 1/(1−b)
⇒ y−by=1
⇒ by=y−1
⇒ b=(y−1)/y
And now,
L.H.S.=1 + ab + a^2b^2 + ....∞
= 1/(1−ab) (infinte G.P. series where (first term)=1 and (common difference)=ab
= 1/{1−(x−1/x)(y−1/y)}
= xy/(xy−xy+x+y−1)
= (xy)/(x+y−1)
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If 1+a+a2+………∞=x and 1+b+b2+…&hellip...
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If 1+a+a2+………∞=x and 1+b+b2+…&hellip...
Explanation:
Given:
1+a+a^2+...∞ = x
1+b+b^2+...∞ = y
To find:
1+ab+a^2b^2+...∞ = ?
Solution:
Step 1:
Let's multiply the two given equations:
(1+a+a^2+...∞)(1+b+b^2+...∞) = x*y
Step 2:
Expanding the left side of the equation:
1 + (a+b) + (a^2 + ab + b^2) + ...∞ = x*y
Step 3:
Now, let's subtract the given equations from the expanded equation:
(x*y) - (x) - (y) = 1 + ab + a^2b^2 + ...∞
Step 4:
Rearranging the terms, we get:
1 + ab + a^2b^2 + ...∞ = x*y - x - y
Step 5:
Now, substitute the values of x and y from the given equations:
1 + ab + a^2b^2 + ...∞ = xy / (x + y - 1)
Therefore, the correct answer is option (A) (xy)/(x+y-1).
Given:
1+a+a^2+...∞ = x
1+b+b^2+...∞ = y
To find:
1+ab+a^2b^2+...∞ = ?
Solution:
Step 1:
Let's multiply the two given equations:
(1+a+a^2+...∞)(1+b+b^2+...∞) = x*y
Step 2:
Expanding the left side of the equation:
1 + (a+b) + (a^2 + ab + b^2) + ...∞ = x*y
Step 3:
Now, let's subtract the given equations from the expanded equation:
(x*y) - (x) - (y) = 1 + ab + a^2b^2 + ...∞
Step 4:
Rearranging the terms, we get:
1 + ab + a^2b^2 + ...∞ = x*y - x - y
Step 5:
Now, substitute the values of x and y from the given equations:
1 + ab + a^2b^2 + ...∞ = xy / (x + y - 1)
Therefore, the correct answer is option (A) (xy)/(x+y-1).
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If 1+a+a2+………∞=x and 1+b+b2+……∞=y then 1+ab+a2b2+………∞ = x is given by ________.a)(xy)/(x+y-1)b)(xy)/(x-y-1)c)(xy)/(x+y+1)d)NoneCorrect answer is option 'A'. Can you explain this answer?
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