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# In the expansion of (k + x)8, the coefficient of x5 is 10 times the coefficient of x6. Find the value of k. - Mathematics and Statistics

Sum

In the expansion of (k + x)8, the coefficient of x5 is 10 times the coefficient of x6. Find the value of k.

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#### Solution

We know that, in the expansion of (a+ b)n,
tr+1 = nCr an–r br

Here, n = 8, a = k, b = x

∴ tr+1 = 8Cr (k)8–r xr          ...(1)

If this is the term containing x5, then r = 5 and Coefficient of x5 = 8C5k3.

Also, if (1) is the term containing x6, then r = 6 and coefficient of x6 = 8C6k2.

Since, coefficient of x5 = 10 × coefficient of x6

8C5k3 = 10 × 8C6k2

∴ (8!)/(5!3!) * "k"^3/"k"^2 = (8!)/(6!2!) xx 10

∴  k = (5!3!)/(6!2!) xx 10

= (5! xx 3 xx 2!)/(6 xx 5! xx 2!) xx 10

∴  k = 5

Concept: General Term in Expansion of (a + b)n
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#### APPEARS IN

Balbharati Mathematics and Statistics 2 (Arts and Science) 11th Standard Maharashtra State Board
Chapter 4 Methods of Induction and Binomial Theorem
Exercise 4.3 | Q 5 | Page 80
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