In the expansion of (32+31)n, n ∈N if the ratio of the 15th term from the beginning to the 15th term from the end is 61 then the value of (3n) is:
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Step-by-Step Solution
Step 1: Identify the terms in the binomial expansion
The general term in the binomial expansion of (a+b)n is given by the formula Tr+1=(rn)an−rbr. In this problem, a=32 and b=31.
Step 2: Express the 15th term from the beginning and the end
The 15th term from the beginning is T15, which corresponds to r=14. The 15th term from the end is T15′. In a binomial expansion of n+1 terms, the k -th term from the end is the (n-k+2) -th term from the beginning. So, the 15th term from the end is the (n−15+2) -th term from the beginning, which is Tn−13. Using the property (kn)=(n−kn), we can write Tn−13 as (n−14n)a14bn−14.
Step 3: Set up the ratio and simplify
We are given that the ratio of the 15th term from the beginning to the 15th term from the end is 61. We substitute the expressions for T15 and T15′ into this ratio. Since (14n)=(n−14n), these terms cancel out.
Step 4: Solve for n
After canceling the binomial coefficients, we combine the terms with the same base. This simplifies to (1/32/3)n−28=61. Further simplification leads to (36)n−28=61. Recognizing that 61=(36)−2, we equate the exponents to find n−28=−2, which gives n=26.
Step 5: Calculate the value of (3n)
Now that we have found n=26, we need to calculate the value of (3n). Substituting n=26, we get (326)=326!!(26−3)!=3×2×126×25×24. Performing the multiplication and division gives the final result.