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Let a=31/203+1 and f(n)= nC0an1 nC1an2+ nC2an3+(1)n1 nCn1a0 where n3. If f(2030)+f(2031)=3x(ya1), then the value of  xy is


Your Answer
A
6
Correct Answer
B
5
Your Answer
C
4
Your Answer
D
4.5

Solution

The correct option is A 5
f(n)= nC0an1 nC1an2            + nC2an3+(1)n1 nCn1a0af(n)= nC0an nC1an1+ nC2an2             +(1)n1 nCn1a             +(1)n nCna0(1)n nCna0af(n)=(1a)n(1)naf(2030)=(1a)20301af(2031)=(1a)2031+1a(f(2030)+f(2031))=(1a)2030+(1a)2031

We know that,
a=31/203+11a=(31/203)
a(f(2030)+f(2031))=(1a)2030(1+(1a))a(f(2030)+f(2031))=310(2a)f(2030)+f(2031)=310×(2a)af(2030)+f(2031)=310×(2a1)3x(ya1)=310×(2a1)x=10,y=2xy=5

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