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Question : 17 of 160
Marks:
+1,
-0
Solution:
Given,
y‌=‌+‌+‌+...+∞‌=‌⋅‌+‌⋅‌+‌⋅‌+...+∞Since, using binomial expansion
(1−x)−n=1+nx+‌x2+‌x3+...Here,
nx=‌‌‌‌⋅⋅⋅⋅⋅⋅⋅(i)‌x2‌=‌⋅‌⇒‌‌n(n+1)x2‌=‌‌‌‌⋅⋅⋅⋅⋅⋅⋅(ii)From Eq. (i) and (ii), we get
‌‌=‌‌=‌=‌⇒‌=‌⇒3n+3=5n⇒n=‌So,
x=‌⋅‌=‌×‌=‌∴‌‌y‌=(1−x)−n−1=(1−‌)−3∕2−1‌=(‌)−3∕2−1=23∕2−1=2√2−1So,
y+1=2√2−1+1=2√2‌(y+1)2=(2√2)2=8⇒y2+2y+1−8=0⇒y2+2y−7=0Thus, the quadratic equation is
y2+2y−7=0
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