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प्रश्न
The coefficient of x2 in the expansion of (1 + 2x)m is 112. Find m
उत्तर
Let tr+1 has x2 in the expansion of (1 + 2x)m.
We know that, in the expansion of (a+ b)n,
tr+1 = nCr an–r br
Here, n = m, a = 1, b = 2x
∴ tr+1 = mCr (1)m–r (2x)r
= mCr2r · xr
But tr+1 has x2
∴ power of x = 2
∴ r = 2
∴ coefficient of x2 = mC2 · 22
But coefficient of x2 is 112
∴ mC2 · 22 = 112
∴ `("m"!)/(2!("m" - 2)!) xx 4` = 112
∴ `("m"("m" - 1)("m" - 2)!)/(2 xx ("m" - 2)!) xx 4` = 112
∴ m(m – 1) = 56 = 8 × 7
∴ m(m – 1) = 8(8 – 1)
By comparing on both sides, m = 8.
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