If the coefficients of (r-5)^{th} and (2r-1)^ | vedclass

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(A) The general term of the expansion $(1+x)^n$ is given by $T_{k+1} = {^nC_k} x^k$. The $(r-5)^{th}$ term is $T_{r-5} = T_{(r-6)+1}$,so its coefficie

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$14$

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(A) The general term of the expansion $(1+x)^n$ is given by $T_{k+1} = {^nC_k} x^k$. The $(r-5)^{th}$ term is $T_{r-5} = T_{(r-6)+1}$,so its coefficient is ${^{34}C_{r-6}}$. The $(2r-1)^{th}$ term is $T_{2r-1} = T_{(2r-2)+1}$,so its coefficient is ${^{34}C_{2r-2}}$. Given that the coefficients are equal,we have ${^{34}C_{r-6}} = {^{34}C_{2r-2}}$. Using the property ${^nC_a} = {^nC_b} \implies a = b$ or $a + b = n$: Case $1$: $r-6 = 2r-2 \implies r = -4$ (Not possible as $r$ must be a natural number). Case $2$: $(r-6) + (2r-2) = 34 \implies 3r - 8 = 34 \implies 3r = 42 \implies r = 14$. Thus,$r = 14$.

If the coefficients of $(r-5)^{th}$ and $(2r-1)^{th}$ terms in the expansion of $(1+x)^{34}$ are equal,find $r$.

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