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NCERT Class XI Mathematics - Binomial Theorem - Solutions

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Question : 22 of 36
Marks: +1, -0
(x3+9y)10\left(\frac{x}{3}+9y\right)^{10}
Solution:  
As the exponent 10 is even, so the middle term of the expansion
[x3+9y]10\left[\frac{x}{3}+9y\right]^{10} is [102+1]\left[\frac{10}{2}+1\right] th term i.e., 6th term which is given by
T6T_6 = T5+1T_{5+1} = 10C5(x3)10−5(9y)5{}^{10}C_5 \left(\frac{x}{3}\right)^{10-5} (9y)^5 = (252) (x3)5(9y)5\left(\frac{x}{3}\right)^5 (9y)^5 = (252) 9535x5y5\frac{9^5}{3^5} x^5 y^5
= (252) 35x5y53^5 x^5 y^5 = 61236 x5y5x^5 y^5
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