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標題:

maths about Binomial Theorem

發問:

If the coefficients of three consecutive terms in the expansion of (1 + x)n are 120 , 210 and 252 respectively, find the value of n .

最佳解答:

Let nCr=120, nC(r+1)=210, nC(r+2)=252 (n!)/(r!(n-r)!)=120 n!=120(r!(n-r)!) ..................................... (1) (n!)/((r+1)!(n-r-1)!)=210 .......................... (2) n!=210((r+1)!(n-r-1)!) Put (1) into (2), (120(r!(n-r)!))/((r+1)!(n-r-1)!)=210 (n-r)/(r+1)=210/120 n-r=(210/120)(r+1) .................................(*) (n!)/((r+2)!(n-r-2)!)=252 n!=252((r+2)!(n-r-2)!) ............................. (3) Put (3) into (2), 252((r+2)!(n-r-2)!)/((r+1)!(n-r-1)!)=210 (r+2)/(n-r-1)=210/252 .............................(**) Put (*) into (**), (r+2)/((210/120)(r+1)-1)=210/252 (-11/24)r=-11/8 r=3 by (*), n-3=(210/120)(3+1) n=10 So, the value of n = 10

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