Binomial theorem for positive integral Index

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Question

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

Moderate

Solution

The general term $T_{r - 1}$ in the expansion of $(3 + 7 x)^{29}$
is
$T_{r + 1} =^{29} C_{r} 3^{29 - r} (7 x)^{r} = \{^{29} C_{r} \times 3^{29 - r} \times 7^{r}\}\} x^{r}$
$∴$ $a_{r}$ = Coefficient of ${(r + 1)}^{t h}$ term $=^{29} C_{r} \times 3^{29 - r} \times 7^{r}$
Now,
$a_{r} = a_{r - 1}$
$\begin{array}{l} \Rightarrow^{29} C_{r} \times 3^{29 - r} \times 7^{r} =^{29} C_{r - 1} \times 3^{30 - r} \times 7^{r - 1} \\ \Rightarrow \frac{^{29} C_{r}}{^{29} C_{r - 1}} = \frac{3}{7} \Rightarrow \frac{30 - r}{r} = \frac{3}{7} \Rightarrow r = 21 \end{array}$

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Binomial theorem for positive integral Index

Remember concepts with our Masterclasses.

If the coefficients of rth and (r+1)th terms in the expansion of (3+7x)29are equal, then r equals

The general term Tr-1 in the expansion of (3+7x)29is Tr+1=29Cr329−r(7x)r= 29Cr×329−r×7rxr∴ ar = Coefficient of (r +1)th term =29Cr×329−r×7rNow, ar=ar−1⇒ 29Cr×329−r×7r=29Cr−1×330−r×7r−1⇒ 29Cr 29Cr−1=37⇒30−rr=37⇒r=21

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