Coefficient of three consecutive terms in the expansion of (1+x)^n are 45,120 and 210 then the value of n is
Coefficient of three consecutive terms in the expansion of (1+x)n are 45,120 and 210 then the value of n is
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Let nC(r-1) = 45, nCr = 120 and nC(r+1) = 210 => nCr / nC(r-1) = 120/45=8/3 and nC(r+1)/nCr = 210/120=7/4 nCr /
nC(r-1)=[ ni /(r)i (n-r)i ]/[ni /(r-1)i (n-r+1)i] = (n-r+1)/r = 8/3 and (n-r)/(r+1) = 7/4 => 4n -11r =7… eq-1 and 3n - 11r =- 3…
eq-2 Subtracting 2nd eqn. from the 1st, n = 10 and r = 3
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