The least value of n for which the number of integral terms in the Binomial expansion of (37+311)n is 183, is:
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Step-by-Step Solution
Step 1: Identify the general term
The general term in the binomial expansion of (a+b)n is given by Tr+1=(rn)an−rbr. In this problem, a=71/3 and b=111/3. We substitute these values into the general term formula.
Step 2: Determine conditions for integral terms
For the term Tr+1 to be an integer, the powers of 7 and 11 must be integers. This means that (n-r) must be a multiple of 3, and r must also be a multiple of 3. Since r is a multiple of 3, n-r being a multiple of 3 implies that n must also be a multiple of 3.
Step 3: Express r in terms of k
Since r must be a multiple of 3, we can write r=3k for some integer k. The possible values for r range from 0 to n. Therefore, 0≤3k≤n, which implies 0≤k≤n/3.
Step 4: Calculate the number of integral terms
The number of integral terms is equal to the number of possible values for k. Since k ranges from 0 to ⌊n/3⌋, the total number of values for k is ⌊n/3⌋+1. We are given that the number of integral terms is 183.
Step 5: Solve for n
We set the expression for the number of integral terms equal to 183 and solve for n. From the definition of the floor function, if ⌊x⌋=m, then m≤x<m+1.
Step 6: Determine the range for n and find the least value
Multiplying the inequality by 3, we find the range for n. Since n must be a multiple of 3 (as established in Step 2), the possible values for n are 546. The least value of n in this range that is a multiple of 3 is 546.