If some three consecutive coefficients in the binomial expansion of (x+1)^n in powers of x are in the ratio 2:15:70, then the average of these three…

From the given condition

$${ }^{n}{C}_{r}{:}^{n}{C}_{r+1}{:}^{n}{C}_{r+2}=2:15:70$$nCr​:nCr+1​:nCr+2​=2:15:70

$\Rightarrow \frac{{ }^{n}{C}_{r}}{{ }^{n}{C}_{r+1}}=\frac{2}{15}$⇒nCr+1​nCr​​=152​ and $\frac{{ }^{n}{C}_{r+1}}{{ }^{n}{C}_{r+2}}=\frac{15}{70}$nCr+2​nCr+1​​=7015​

$\Rightarrow \frac{ (\frac{n!}{(n-r)!\cdot r!})}{(\frac{n!}{(n-r-1)!\cdot (r+1)!})}=\frac{2}{15}$⇒((n−r−1)!⋅(r+1)!n!​)((n−r)!⋅r!n!​)​=152​ and $\frac{ (\frac{n!}{(n-r-1)!\cdot (r+1)!})}{(\frac{n!}{(n-r-2)!\cdot (r+2)!})}=\frac{15}{70}$((n−r−2)!⋅(r+2)!n!​)((n−r−1)!⋅(r+1)!n!​)​=7015​

$\Rightarrow \frac{(n-r-1)!\cdot (r+1)!}{(n-r)!\cdot r!}=\frac{2}{15}$⇒(n−r)!⋅r!(n−r−1)!⋅(r+1)!​=152​ and $\frac{(n-r-2)!\cdot (r+2)!}{(n-r-1)!\cdot (r+1)!}=\frac{3}{14}$(n−r−1)!⋅(r+1)!(n−r−2)!⋅(r+2)!​=143​

$\Rightarrow \frac{(n-r-1)!\cdot (r+1)\cdot r!}{(n-r)\cdot (n-r-1)!\cdot r!}=\frac{2}{15}$⇒(n−r)⋅(n−r−1)!⋅r!(n−r−1)!⋅(r+1)⋅r!​=152​ and $\frac{(n-r-2)!\cdot (r+2)\cdot (r+1)!}{(n-r-1)\cdot (n-r-2)!\cdot (r+1)!}=\frac{3}{14}$(n−r−1)⋅(n−r−2)!⋅(r+1)!(n−r−2)!⋅(r+2)⋅(r+1)!​=143​

$\Rightarrow \frac{ r+1}{n-r}=\frac{2}{15}$⇒n−rr+1​=152​ and $\frac{ r+2}{n-r-1}=\frac{3}{14}$n−r−1r+2​=143​

$\Rightarrow 17r=2n-15$⇒17r=2n−15 and $17r=3n-31$17r=3n−31

$\Rightarrow 3n-31=2n-15,\Rightarrow n=16$⇒3n−31=2n−15,⇒n=16 and $r=1$r=1

Hence, average $=\frac{{ }^{n}{C}_{r}{+}^{n}{C}_{r+1}{+}^{n}{C}_{r+2}}{3}$=3nCr​+nCr+1​+nCr+2​​

$$=\frac{{ }^{16}{C}_{1}{+}^{16}{C}_{2}{+}^{16}{C}_{3}}{3}$$=316C1​+16C2​+16C3​​

$$=232.$$=232.