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
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
⇒|nCrxn−ryrnCr−1xn−r+1yr−1|≥1
⇒n−r+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]
5≤n+12.5≤6
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