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Find the coefficient of x4 in the expansion (1+x3)50(x2+1x)5 - Mathematics

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प्रश्न

Find the coefficient of x4 in the expansion `(1 + x^3)^50 (x^2 + 1/x)^5`

बेरीज

उत्तर

`(x^2 + 1/x)^5 = ""^5"C"_0 (x^2)^5 + ""^5"C"_1 (x^2)^4 (1/x) + ""^5"C"_2 (x2)^3 (1/x)^2 + ""^5"C"_3 (x^2)^2 (1/x)^3 + ""^5"C"_4 (x^2) (1/x)^4 + ""^5"C"_5 (1/x)^5`

5C0 = 1 = 5C5 ; 5C1 = 5 = 5C4 ; 5C2 = `(5 xx 4)/(2 xx 1)` = 10 = 5C3

= `x^10 + 5(x^8) (1/x) + 10(x^6) (1/x^2) + 10(x^4) (1/x^3)+ 5(x^2) (1/x^4) + 1/x^5` 

= `x^10 + 5x^7 + 10x^4 + 10x + 5/x^2 + 1/x^5`

(1 + x3)50 = `""^50"C"_0 (1)^50 (x^3)^0 + ""^50"C"_1 (1)^49 (x^3)^1 + ""^50"C"_2 (1)^48 (x^3)^2 +""^50"C"_3 (1)^7 (x^3)^3 + .... ""^50"C"_50 (1)^circ(x^3)^50`

50C0 = 1 = 50C50, 50C1 = 50, 50C2 = 1225, 50C3 = 19600

= (50) + (50)x3 + 1225x6 7600x9.... x150

To find coefficient of x4

`(1 + x^3)^50 (x^2 + 1/x)^5 = (1 + 50x^3 + 1225x^6 + 19600x^9 ...  x^150) xx (x^10 + 5x^7 + 10x^4 + 10x^4 + 10x + 5/x^2 + 1/x^5)`

When multiplying these terms, we get x4 terms

= `(1 xx 10x^4) + (50x^3 xx 10x) + (1225x^6 xx5/x^2) + (19600x^9 xx 1/x^5)`

= 10x4 + 500x4 + 6125x4 + 19600x4

= 26325x4

∴ The coefficient of x4 is 26325

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Binomial Theorem
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पाठ 5: Binomial Theorem, Sequences and Series - Exercise 5.1 [पृष्ठ २१०]

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सामाचीर कलवी Mathematics - Volume 1 and 2 [English] Class 11 TN Board
पाठ 5 Binomial Theorem, Sequences and Series
Exercise 5.1 | Q 6 | पृष्ठ २१०
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