Tr+1=12Cr(x31/5)12−r(51/32x)rTr+1=(5)r/312Cr(3)512−r(2)r(x)2r−12r=6 T7=5212C6(3)6/5(2)6=(259×11×7)28⋅31/525α=693
If the constant term in the expansion of ( √[5]3 x+ 2 x √[3]5 )^12, x ≠ 0, is α × 2^8 × √[5]3, then 25 α is equal to :
JEE Main 2024 — Mathematics Algebra
If the constant term in the expansion of (x53+352x)12,x=0, is α×28×53, then 25α is equal to :
Held on 5 Apr 2024 · Verified 6 Jul 2026.
Options
- A
724
- B
742
- C
639
- D
693
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