The coefficients of three successive terms in | vedclass

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Question

(A) Let the coefficients of three consecutive terms,i.e.,$(r+1)^{th}, (r+2)^{th},$ and $(r+3)^{th}$ in the expansion of $(1+x)^n$ be $^nC_r = 165$,$^n

Vedclass · Accepted Answer

$11$

Vedclass · Answer

(A) Let the coefficients of three consecutive terms,i.e.,$(r+1)^{th}, (r+2)^{th},$ and $(r+3)^{th}$ in the expansion of $(1+x)^n$ be $^nC_r = 165$,$^nC_{r+1} = 330$,and $^nC_{r+2} = 462$.Using the ratio formula $\frac{^nC_{r+1}}{^nC_r} = \frac{n-r}{r+1}$,we get:$\frac{330}{165} = \frac{n-r}{r+1} \implies 2 = \frac{n-r}{r+1} \implies n-r = 2r+2 \implies n = 3r+2$ (Equation $1$).Using the ratio formula $\frac{^nC_{r+2}}{^nC_{r+1}} = \frac{n-(r+1)}{r+2}$,we get:$\frac{462}{330} = \frac{n-r-1}{r+2} \implies \frac{7}{5} = \frac{n-r-1}{r+2}$.Cross-multiplying gives $7(r+2) = 5(n-r-1) \implies 7r+14 = 5n-5r-5 \implies 5n = 12r+19$ (Equation $2$).Substituting $n = 3r+2$ from Equation $1$ into Equation $2$:$5(3r+2) = 12r+19 \implies 15r+10 = 12r+19 \implies 3r = 9 \implies r = 3$.Now,substituting $r=3$ into Equation $1$:$n = 3(3)+2 = 11$.Thus,the value of $n$ is $11$.

The coefficients of three successive terms in the expansion of $(1 + x)^n$ are $165, 330$ and $462$ respectively. Then the value of $n$ is:

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