Binomial theorem for positive integral Index

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Question

If the coefficients of the ${(r + 2)}^{t h}$ and $r^{t h}, {(r + 1)}^{t h}$ terms in the binomial expansion of ${(1 + y)}^{m}$ are in , then $A . P .$ $m$ and $r$ satisfy the equation

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Solution

Coefficient of the $r th$ term is $^{m} C_{r - 1}$
According to the given condition $n^{m} C_{r - 1} +^{m} C_{r + 1} = 2 (^{m} C_{r})$
$\begin{array}{ll} \Rightarrow \frac{^{m} C_{r - 1}}{^{m} C_{r}} + \frac{^{m} C_{r + 1}}{^{m} C_{r}} = 2 \Rightarrow \frac{r}{m + 1 - r} + \frac{m - r}{r + 1} = 2 \\ \Rightarrow r^{2} + r + (m - r) (m + 1 - r) = 2 (m + 1 - r) (r + 1) \\ \Rightarrow r^{2} + r + m^{2} + (1 - 2 r) m - r (1 - r) = \\ \Rightarrow 2 m (r + 1) + 2 (1 - r^{2}) \\ \Rightarrow m^{2} - m (4 r + 1) + 4 r^{2} - 2 = 0 . \end{array}$

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Binomial theorem for positive integral Index

Remember concepts with our Masterclasses.

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

Coefficient of the r th term is mCr−1According to the given condition nmCr−1+mCr+1=2 mCr⇒ mCr−1 mCr+ mCr+1 mCr=2⇒rm+1−r+m−rr+1=2⇒ r2+r+(m−r)(m+1−r)=2(m+1−r)(r+1)⇒ r2+r+m2+(1−2r)m−r(1−r)= ⇒2m(r+1)+21−r2⇒ m2−m(4r+1)+4r2−2=0.

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If the coefficients of the ${(r + 2)}^{t h}$ and $r^{t h}, {(r + 1)}^{t h}$ terms in the binomial expansion of ${(1 + y)}^{m}$ are in , then $A . P .$ $m$ and $r$ satisfy the equation