If the seventh terms from the beginning and the end in the expansion of (23+133)n are equal, then n equals ______. - Mathematics

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If the seventh terms from the beginning and the end in the expansion of (root(3)(2) + 1/(root(3)(3)))^n are equal, then n equals ______.

Solution

If the seventh terms from the beginning and the end in the expansion of (root(3)(2) + 1/(root(3)(3)))^n are equal, then n equals 12.

Explanation:

The given expansion is (root(3)(2) + 1/(root(3)(3)))^n

∴ T7 = T6+1

= ""^n"C"_6 (2^(1/3))^(n - 6) * 1/((3^(1/3))^6

= ""^n"C"_6 (2)^((n - 6)/3) * 1/(3)^2

Now the T7 from the end = T7  from the beginning in (1/(root(3)(2)) + root(3)(2))^n.

∴ T7 = T6+1

= ""^n"C"_6 (1/(3^(1/3)))^(n - 6) * (2^(1/3))^6

We get ""^n"C"_6 (2)^((n - 6)/3) * (1/3^2) = ""^n"C"_6 1/((n - 6)/3) * (2)^2

⇒ (2)^((n - 6)/3) * (3)^-2 = (3)^-((n - 5)/3) * (2)^2

⇒ (2)^((n - 6)/3  2)  * (3)^(-2 + (n - 6)/3) = 1

⇒ 2^((n - 12)/3) * (3)^((n - 12)/3) = 1

⇒ (6)^((n - 12)/3) = (6)^0

⇒ (n - 12)/3 = 0

⇒ n = 12

Concept: Rth Term from End
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APPEARS IN

NCERT Mathematics Exemplar Class 11
Chapter 8 Binomial Theorem
Exercise | Q 28 | Page 145
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