If the coefficients ofn r^th ,(r + 1)^th andn (r + 2)^th terms in the binomial expansion ofn (1 + y)^m are in A.P., then m and r satisfy then equation

(b) : T_r+1= ^m C_r y^rn nn n Therefore ^m C_T-1+ _m!^m C_T+1=2 x ^m C_T n => (m! / (r-1)!(m-r+1)!)+(m! / (r+1)!(m-r-1)!)=2 (m! / r!(m-r)!) n => (r(r+1) / (r+1)!(m-r+1)!)+((m-r+1)(m-r) / (r+1)!(m-r+1)!) n =(2(r+1)(m-r+1) / (r+1)!(m-r+1)!) n => r(r+1)+(m-r+1)(m-r)=2(r+1)(m-r+1) n => r r(r+1)+(m-r)^2+m-r-2(r+1)(m-(r-1))=0 n => r(r+1)+m^2+r^2-2 m r+m-r+2(r^2-1)-2 m(r+1)=0 n => m^2-m(4 r+1)+4 r^2-2=0 \ (i)n n

If the coefficients of r^{{th}{~}},(r + 1)^{{th}{~}} ... | Sequence & Series (Mathematics)

If the coefficients of\n

r^{\text{th}\text{~}},(r + 1)^{\text{th}\text{~}}

and\n

(r + 2)^{\text{th}\text{~}}

terms in the binomial expansion of\n

(1 + y)^{m}

are in A.P., then

m

and

r

satisfy the\n equation

m^{2} - m(4r - 1) + 4r^{2} + 2 = 0

m^{2} - m(4r + 1) + 4r^{2} - 2 = 0

m^{2} - m(4r + 1) + 4r^{2} + 2 = 0

m^{2} - m(4r - 1) + 4r^{2} - 2 = 0

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Correct Answer:

m^{2} - m(4r + 1) + 4r^{2} - 2 = 0

Explanation

(b)

: T_{r+1}={ }^m C_r y^r

\n \n

\n\begin{aligned}\n& \therefore{ }^m C_{T-1}+{ }_{m!}^m C_{T+1}=2 \times{ }^m C_T \\\n& \Rightarrow \frac{m!}{(r-1)!(m-r+1)!}+\frac{m!}{(r+1)!(m-r-1)!}=2 \frac{m!}{r!(m-r)!} \\\n& \Rightarrow \frac{r(r+1)}{(r+1)!(m-r+1)!}+\frac{(m-r+1)(m-r)}{(r+1)!(m-r+1)!} \\\n& =\frac{2(r+1)(m-r+1)}{(r+1)!(m-r+1)!} \\\n& \Rightarrow r(r+1)+(m-r+1)(m-r)=2(r+1)(m-r+1) \\\n& \Rightarrow r r(r+1)+(m-r)^2+m-r-2(r+1)(m-(r-1))=0 \\\n& \Rightarrow r(r+1)+m^2+r^2-2 m r+m-r+2\left(r^2-1\right)-2 m(r+1)=0 \\\n& \quad \Rightarrow m^2-m(4 r+1)+4 r^2-2=0 \#(i)\n\end{aligned}\n

Explanation

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