Question

# If the seventh term from the beginning and end in the binomial expansion of (3√2+13√3)n are equal, find n.

Open in App
Solution

## Given expansions is (3√2+13√3)n ∴T7=T6+1=nC6(3√2)n−6(13√3)6 ...(i) Since, T7 from end is same as the T7 from beginning of (13√3+3√2)2 Then, T7=nC6(13√3)n−6(3√2)6 ...(ii) Given that, nC6(2)n−63(3)−62=nC6(3)(n−6)3(2)63 ⇒(2)n−123=(1313)n−12 which is true, when n−123=0 ⇒n−12=0⇒n=12

Suggest Corrections
4
Explore more