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Question

If 6th term in the expansion of (32+x3)n is the numerically greatest term when x=3, then find the sum of all possible values of n


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Solution

In the binomial expansion of (x+y)n, if we put certain values of a and b, each term will have certain value. The term which has the numerically greatest value is called the numerically greatest term. Let us assume Tr+1 has the numerically greatest term. Then, |Tr+1||Tr|
|Tr+1Tr|1

|nCrxnryrnCr1xnr+1yr1|1

nr+1r|yx|1

When we solve for r, we get

r=[(n+1)(1+|xy|)]

T6 is the numerically greatest term.
r=5
Also r=[(n+1)(1+|xy|)]

x and y in the equationresult is different from the one in question
5=[n+1(1+3×32×x)]

5=[n+1)(1+3×32×3)]

5=[n+12.5]

5n+12.56
12.5 n+1 < 15
11.5 n < 14
n is an integer and it can take the values 12 and 13
Sum of the possible values =12+13=25


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