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r and n are positive integers r> 1,n> 2 and coefficient of (r+2)^{th} term and 3r^{th} term in the expansion of  (1+x)^{2n} are equal, then n equals

  • Option 1)

    3r\;

  • Option 2)

    \; 3r+1\;

  • Option 3)

    \; 2r\;

  • Option 4)

    \; 2r+1

 

Answers (1)

As we learnt in

Properties of Binomial Theorem -

If ^{n}c_{r}= ^{n}c_{s} then

r= s \, or \, r+s= n

-

 

 In (1+x)^{2n}

Coefficient of T_{r+2}=\ ^{2n}C_{r+1}  and Coefficient of T_{3r}=\ ^{2n}C_{3r-1}

So if \ ^{2n}C_{r+1}=\ ^{2n}C_{3r-1}

Then either r + 1 = 3r - 1 

i.e. r = 1

or r + 1 + 3r - 1 = 2n

r=\frac{n}{2}\ \; \Rightarrow\ \;n = 2r

Correct option is 3.

 


Option 1)

3r\;

This is an incorrect option.

Option 2)

\; 3r+1\;

This is an incorrect option.

Option 3)

\; 2r\;

This is an incorrect option.

Option 4)

\; 2r+1

This is an incorrect option.

Posted by

Sabhrant Ambastha

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