JEE maths#12

Probability Level 3

( 1 − 2 x + 5 x 2 + 10 x 3 ) ( 1 + x ) n = 1 + a 1 x + a 2 x 2 + ⋯ (1-2x + 5x^{2} + 10x^{3})(1+x)^{n} = 1+a_{1}x + a_{2}x^{2}+ \cdots and ( a 1 ) 2 = 2 a 2 (a_{1})^{2} = 2a_{2}

Then find the value of n n .

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The answer is 6.

Rahil Sehgal by Rahil Sehgal , 20, India

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1 solution

Tapas Mazumdar
Apr 10, 2017

( 1 − 2 x + 5 x 2 + 10 x 3 ) ( 1 + x ) n = ( 1 − 2 x + 5 x 2 + 10 x 3 ) [ ( n 0 ) + ( n 1 ) x + ( n 2 ) x 2 + ⋯ + ( n n ) x n ] = 1 + [ − 2 ( n 0 ) + ( n 1 ) ] x + [ 5 ( n 0 ) − 2 ( n 1 ) + ( n 2 ) ] x 2 + ⋯ = 1 + [ n − 2 ] ⏟ = a 1 x + [ n ( n − 1 ) 2 − 2 n + 5 ] ⏟ = a 2 x 2 + ⋯ \begin{aligned} (1-2x+5x^2+10x^3) (1+x)^n &= (1-2x+5x^2+10x^3) \left[ \dbinom n0 + \dbinom n1 x + \dbinom n2 x^2 + \cdots + \dbinom nn x^n \right] \\ &= 1 + \left[ -2 \dbinom n0 + \dbinom n1 \right] x + \left[ 5 \dbinom n0 - 2 \dbinom n1 + \dbinom n2 \right] x^2 + \cdots \\ &= 1+ \underbrace{\left[ n-2 \right]}_{= \ a_1} x + \underbrace{\left[ \dfrac{n(n-1)}{2} -2n+5 \right]}_{= \ a_2} x^2 + \cdots \end{aligned}

Hence

( a 1 ) 2 = 2 a 2 ⟹ ( n − 2 ) 2 = 2 [ n ( n − 1 ) 2 − 2 n + 5 ] ⟹ n = 6 (a_1)^2 = 2a_2 \implies (n-2)^2 = 2 \left[ \dfrac{n(n-1)}{2} -2n+5 \right] \implies n = \boxed{6}

1 Helpful 0 Interesting 0 Brilliant 0 Confused

Thank you very much (+1)

Rahil Sehgal - 4 years, 2 months ago

@Calvin Lin Sir please provide an appropriate rating for the problem...

Thank you... :)

Rahil Sehgal - 4 years, 1 month ago

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I have set a seed level for the problem.

In future, if you set it when you first create the problem, that will allow us to accumulate data right from the start, especially since that is when most problems get their initial attention.

Calvin Lin Staff - 4 years, 1 month ago

There is one more solution which involve differentiating the given equation twice and then substituting the value in the given constant

Aniket C-mmon - 3 years, 1 month ago

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