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