Binomial Theorem

Question
CBSEENMA11015613

For natural numbers m, n if (1 − y)m (1 + y)n = 1 + a1y + a2y2 + … , and a1 = a2 = 10, then (m, n) is

  • (20, 45)

  • (35, 20)

  • (45, 35)

  • (35, 45)

Solution

D.

(35, 45)

left parenthesis 1 minus straight y right parenthesis to the power of straight m space left parenthesis 1 plus straight y right parenthesis to the power of straight n space equals space left square bracket 1 minus to the power of straight m straight C subscript 1 straight y space plus to the power of straight m straight C subscript 2 straight y squared space minus........ right square bracket left square bracket 1 plus to the power of straight n straight C subscript 1 straight y space plus to the power of straight n straight C subscript 2 straight y squared plus... right square bracket
space equals space 1 space plus space left parenthesis straight n minus straight m right parenthesis space plus open curly brackets fraction numerator straight m left parenthesis straight m minus 1 right parenthesis over denominator 2 end fraction plus fraction numerator straight n left parenthesis straight n minus 1 right parenthesis 2 over denominator 2 end fraction minus mn close curly brackets straight y squared space plus....
therefore space straight a subscript 1 space equals straight n minus straight m space equals space 10 space and space straight a subscript 2 space equals space fraction numerator straight m squared plus straight n squared space minus straight m minus straight n minus 2 mn over denominator 2 end fraction space equals space 10
So comma space straight n minus straight m equals 10 space and space left parenthesis straight m minus straight n right parenthesis squared space minus left parenthesis straight m plus straight n right parenthesis space equals space 20
rightwards double arrow straight m plus straight n space equals space 80
therefore comma space straight m space equals space 35 comma space straight n space equals space 45

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