If the second, third and fourth terms in the expansion of (x+y )^n are 135,30 and 103, respectively, then 6(n^3+x^2+y) is equal to______

Question

Accepted Answer

Correct answer: 806. (806) ^nC_1x^n-1y=135 … (i ) ^nC_2x^n-2y^2=30 . … . ( ii ) ^nC_3x^n-3y^3= 103 … . ( iii ) By (i ) ( ii ) ^nC_1 ^nC_2 xy= 92 … ( iv ) By ( ii ) ( iii ) ^nC_2 ^nC_3 xy=9 … (v ) By ⁡ ( iv ) (v ) ^nC_1 ^nC_3 ^nC_2 ^nC_2= 12 2n^2 (n-1 ) (n-2 )6= n (n-1 )2 n (n-1 )24n-8=3n-3 ⇒n=5 put in (v ) xy=9x=9y put in (i ) ^5C_1x^4 ( x9 )=135x^5=27 9 ⇒x=3 , y= 136(n^3+x^2+y) =6(125+9+ 13) =806

$If the second, third and fourth terms in the expansion of (x + y)^{n} are 135,30 and \frac{10}{3}, respectively, then 6 (n^{3} + x^{2} + y) is equal to______$

Solution

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