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The largest term in the expansion of (3+...

The largest term in the expansion of `(3+2x)^50`, where x=1/5, is

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The correct Answer is:
`T_(6) = 3^(50) xx .^(50)C_(5)((2)/(15))^(5)` or `T_(7) = 3^(50) xx .^(50)C_(6)(2/15)^(6)`
We have
`(3+2x)^(50)`
Now, `(T_(r+1))/(T_(r)) = (50-r+1)/(r ) . (2x)/(3)`
`rArr (T_(r+1))/(T_(r)) = (51-r)/(r ) .(2x)/(3)` (Putting `x = 1//5`)
Let `(T_(r+1))/(T_(r)) ge 1`
`rArr 102 - 2r ge 15r`
`rArr 102 ge 17r`
`rArr r le 6`
Hence, `7^(th)` and `6^(th)` terms are the greatest.
Now, `T_(7) = T_(6+1) = 3^(50) xx .^(50)C_(6)(2/15)^(6)`
`T_(6) = T_(5+1) = 3^(50) xx .^(50)C_(5) (2/15)^(5)`
`:. (T_(7))/(T_(6)) = (.^(50)C_(6))/(.^(50)C_(5)) (2)/(15)`
`= (50-6+1)/(6) = 2/15`
`= 45/6 2/15 = 1`
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