हिंदी

Find the term independent of x in the expansion of (x3+32x2)10.

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प्रश्न

Find the term independent of x in the expansion of `(sqrt(x)/sqrt(3) + sqrt(3)/(2x^2))^10`.

योग
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उत्तर

Let (r + 1)th term be independent of x which is given by

Tr+1 = `""^10"C"_r  sqrt(x/3)^(10 - r)  sqrt(3)^r/(2x^2)`

= `""^10"C"_r  x^((10 - r)/2)/3  3^(r/2)  1/(2^r  x^(2r))`

= `""^10"C"_r  3^(r/2 - (10 - r)/2)  2^(-r)  x^((10 - r)/2 - 2r)` 

Since the term is independent of x, we have

`(10 - r)/2 - 2r` = 0

⇒ r = 2

Hence 3rd term is independent of x and its value is given by

T3 = `""^10"C"_2  (3^(-3))/4`

= `(10 xx 9)/(2 xx 1) xx 1/(9 xx 12)`

= `5/12`

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अध्याय 8: Binomial Theorem - Solved Examples [पृष्ठ १३४]

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एनसीईआरटी एक्झांप्लर Mathematics [English] Class 11
अध्याय 8 Binomial Theorem
Solved Examples | Q 7 | पृष्ठ १३४

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