Quadratic Equations Question - None

Question

Let $\sum_{r=1}^{10}(r+r \times \binom{10}{r}) = 2^{10}(a \times 4^5 + b)$ where $a, b \in \mathbb{N}$ and $f(x) = x^2 - 2x - k^2 + 1$.

If a and b lie between the roots of $f(x) = 0$, then find the smallest positive integral value of k.

Accepted Answer

Step 1: Evaluate the sum: $\sum_{r=1}^{10}(r + r imes \binom{10}{r}) = \sum_{r=1}^{10}r(1+\binom{10}{r})$.

Step 2: Using identities: $\sum_{r=0}^{10}r\binom{10}{r} = 10 imes 2^9 = 5 imes 2^{10}$ and $\sum_{r=1}^{10}r = 55$.

Step 3: This gives $2^{10}(5 imes 4 + 0.0546875)$, so $a = 5, b = 1$.

Step 4: For a = 5 and b = 1 to lie between roots of $f(x) = x^2 - 2x - k^2 + 1 = 0$, we need $f(1) < 0$ and $f(5) < 0$.

Step 5: $f(1) = 1 - 2 - k^2 + 1 = -k^2 < 0$ (always true for $k > 0$).

$f(5) = 25 - 10 - k^2 + 1 = 16 - k^2 < 0 \Rightarrow k^2 > 16 \Rightarrow k > 4$.

∴ The smallest positive integral value of k is 5.

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