For natural number m,n, if (1−y)m(1−y)n=1+a1y+a2y2+........ and a1=a2=10, then (m,n) is
A
(45, 35)
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B
(35, 45)
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C
(20, 45)
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D
(35, 20)
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Solution
The correct option is C (35, 45) (1−y)m(1+y)n=1+a1y+a2y2+....... ⇒(1−mC1y+mC2y2−...........)(1+nC1y+nC2y2+.......)=1+a1y+a2y2+........... ⇒a1=nC1−mC1,a2=mC2−mCn1C1+nC2 ⇒n−m=10,m2+n2−m−n−2mn=20 ⇒n−m=10,(m−n)2−(m+n)=20 ⇒n−m=10,n+m=80 ⇒m=35,n=45