binomial theorem question

We can use simple expansion up to Degree 2 to find the value of n.


( 2 - x )^2 ( 1 + 2x )^n

= ( 4 - 4x + x^2 )[ 1 + nC1( 2x ) + nC2( 2x )^2 + .... ]

= ( x^2 - 4x + 4 )[ 2n( n - 1 )x^2 + 2nx + 1 + ..... ]

= [ 1 - 8n + 8n( n - 1 ) ]x^2 + ( 8n - 4 )x ......

= ( 8n^2 - 16n + 1 )x^2 + ( 8n - 4 )x ......

From the ratio of coefficients of x and x^2, we have

28( 8n^2 - 16n + 1 ) = 65( 8n - 4 )

224n^2 - 968n + 288 = 0

28n^2 - 121n + 36 = 0

( n - 4 )( 28n - 9 ) = 0

n = 4 or n = 9/28 ( rejected as n is a positive integer )

Hence, n = 4
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Hope I can help you.

(2 - x)^2 (1 + 2x)^n

= (4 - 4x + x^2)(1 + 2nx + 2n(n-1)x^2 + ...)

= 4 + 8nx + 8n(n - 1)x^2 - 4x - 8nx^2 + x^2

= 4 + (8n - 4)x + (8n(n - 1) - 8n + 1)x^2

So, 8n - 4 : 8n^2 - 16n + 1 = 28 : 65

n = 4