TS ICET 23 May 2018 Shift 2 Paper

Question : 131 of 200

Marks: +1, -0

If the coefficients of

x^7

and

x^8

are equal in the expansion of

(3+x/2)^n

then n =

Solution:

The coefficient of

x^7

(3+\;{x}/{2})^{2}

^{n} C_{7} 3^{n-7}(\;{1}/{2})^{7}

Given,

^{n} C_{7} 3^{n-7}(\;{1}/{2})^{7}

The coefficient of

x^8

(3+\;{x}/{2})^{n}

^{n} C_{8} 3^{n-8}(\;{1}/{2})^{8}

=^{n} C_{8} 3^{n-8}({1}/{2})^{8}

⇒{n !}/{(n-7) ! 7 !}{3^{n}}/{3^{7}} ⋅{1}/{2^{7}}=\;{n !}/{(n-8) ! 8 !} {3^{n}}/{3^{8}} ⋅ {1}/{2^{8}}

=\;{1}/{n-7}=\;{1}/{8.3 .2}

⇒ n-7=48

⇒ n=55

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