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If the coefficients of x7 & x8 in the expansion of [2+x/3]n are equal then the value of n is:

A 15
B 45
C 55
D 56

1 Answer

6 votes

Final answer:

The value of n for which the coefficients of x^7 and x^8 in the expansion of (2 + x/3)^n are equal is found using the binomial theorem. Simplifying the equation derived from setting the two coefficients equal, we find that n must be 56.

Step-by-step explanation:

To find the value of n for which the coefficients of x7 and x8 in the expansion of (2 + x/3)n are equal, we use the binomial theorem.

The binomial theorem states that the coefficient of xk in the expansion of (a + b)n is given by nCk × an-k × bk.

Thus, the coefficients we are looking for are nC7 × 2n-7 × (1/3)7 and nC8 × 2n-8 × (1/3)8 respectively.

To have these coefficients equal, we set the two expressions equal to each other and simplify:

  • nC7 × 2n-7 × (1/3)7 = nC8 × 2n-8 × (1/3)8

By canceling common terms and simplifying, we find that n must equal 56.

Therefore, the answer is D: 56.

User Sander Aernouts
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