If coefficients of \(\mathrm{x}^{7}\) and \(\mathrm{x}^{8}\) are equal in expansion of \([2+(\mathrm{x} / 3)]^{\mathrm{n}}\) then \(\mathrm{n}=\) (a) 55 (b) 56 (c) 54 (d) 58

The binomial theorem states that the expansion of (a + b)^n, where n is a non-negative integer, can be written as: \((a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k}b^k\) Where \(\binom{n}{k} = \frac{n!}{k!(n-k)!}\) are the binomial coefficients. In our case, a = 2 and b = x/3.

Using the binomial theorem, we find the coefficients of x^7 and x^8 in the expansion: Coefficient of x^7: \( \binom{n}{k} (2)^{n-k}(x/3)^{k} = x^7\) \( \binom{n}{k} (2)^{n-k}(x/3)^{7} = x^7\) Here, k = 7. Similarly, the coefficient of x^8: \( \binom{n}{k} (2)^{n-k}(x/3)^k = x^8\) \( \binom{n}{k} (2)^{n-k}(x/3)^8 = x^8\) Here, k = 8.

Since the coefficients of x^7 and x^8 are equal, we can equate the expressions we found in step 2: \(\frac{n!}{7!(n-7)!} \cdot (2)^{n-7}\cdot(\frac{1}{3^7}) = \frac{n!}{8!(n-8)!}\cdot (2)^{n-8}\cdot(\frac{1}{3^8})\) Now, we solve for n: \(\frac{8}{3^1} \cdot \frac{(n-7)!}{(n-8)!} = 1\) \(8 = 3(n-7)\) \(n - 7 = \frac{8}{3}\) \(n = 15\) Since n is an integer and 15 does not match any of the given options, we must further analyze the problem.

Notice that in (a), (b), and (d), the value of n is approximately doubled. Upon closer inspection, we can see that if we double the value of n we obtained in step 3, we get: n = 15 * 2 = 30 Let's plug this into our equation and see if it works: \[ \frac{30!}{7!(30-7)!} \cdot (2)^{30-7}\cdot(\frac{1}{3^7}) = \frac{30!}{8!(30-8)!} \cdot (2)^{30-8}\cdot(\frac{1}{3^8}) \] Now it's only a matter of simplification: \[ \frac{30!}{7!(23)!} \cdot (2)^{23}\cdot(\frac{1}{3^7}) = \frac{30!}{8!(22)!} \cdot (2)^{22}\cdot(\frac{1}{3^8})\] Dividing both sides by \(\frac{30!}{3^7 \cdot 2^{22}}\), we get: \[ \frac{1}{7! \cdot 23!} = \frac{1}{8! \cdot 22!}\] Which is true. Thus, when we adjust the result from step 3 to n = 30, it still holds true that the coefficients of x^7 and x^8 are equal. Therefore, none of the given options are correct.