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Question -

Find the 11th term from the beginning and the 11th termfrom the end in the expansion of (2x – 1/x2)25.



Answer -

Given:

(2x – 1/x2)25

The given expressioncontains 26 terms.

So, the 11th termfrom the end is the (26 − 11 + 1) th term from thebeginning.

In other words, the 11th termfrom the end is the 16th term from the beginning.

Then,

T16 =T15+1 = 25C15 (2x)25-15 (-1/x2)15

25C15 (210)(x)10 (-1/x30)

= – 25C15 (210 /x20)

Now we shall find the11th term from the beginning.

T11 =T10+1 = 25C10 (2x)25-10 (-1/x2)10

25C10 (215)(x)15 (1/x20)

25C10 (215 /x5)

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