If the coefficients of r th and (r+1) th term | vedclass

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(D) The general term in the expansion of $(a+bx)^n$ is $T_{k+1} = {^nC_k} a^{n-k} (bx)^k$. For the expansion of $(3+7x)^{29}$,the $r$ th term is $T_r

Vedclass · Accepted Answer

$21$

Vedclass · Answer

(D) The general term in the expansion of $(a+bx)^n$ is $T_{k+1} = {^nC_k} a^{n-k} (bx)^k$. For the expansion of $(3+7x)^{29}$,the $r$ th term is $T_r = {^{29}C_{r-1}} (3)^{29-(r-1)} (7x)^{r-1}$. The coefficient of the $r$ th term is ${^{29}C_{r-1}} (3)^{30-r} (7)^{r-1}$. The $(r+1)$ th term is $T_{r+1} = {^{29}C_r} (3)^{29-r} (7x)^r$. The coefficient of the $(r+1)$ th term is ${^{29}C_r} (3)^{29-r} (7)^r$. Given that these coefficients are equal: ${^{29}C_{r-1}} (3)^{30-r} (7)^{r-1} = {^{29}C_r} (3)^{29-r} (7)^r$ Divide both sides by ${^{29}C_{r-1}} (3)^{29-r} (7)^{r-1}$: $3 = {^{29}C_r} / {^{29}C_{r-1}} 	imes 7$ $3/7 = {^{29}C_r} / {^{29}C_{r-1}}$ Using the property ${^nC_r} / {^nC_{r-1}} = (n-r+1)/r$: $3/7 = (29-r+1) / r$ $3/7 = (30-r) / r$ $3r = 7(30-r)$ $3r = 210 - 7r$ $10r = 210$ $r = 21$

If the coefficients of $r$ th and $(r+1)$ th terms in the expansion of $(3+7x)^{29}$ are equal,then $r$ is equal to

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