Question
Let the smallest value of $k \in \mathbb{N}$, for which the coefficient of $x^{3}$ in $(1 + x)^{3} + (1 + x)^{4} + (1 + x)^{5} + \ldots + (1 + x)^{99} + (1 + kx)^{100},x \neq 0$, is $\left( 43n + \frac{101}{4} \right)$ ( $\ ^{100}C_{3}$ ) for some $n \in \mathbb{N}$, be $p$. Then the value of $p$ +n is:
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Solution
Sol. [Key Concept: The series telescopes (hockey-stick) to $C(100,4)$; match the given form for $n$ and $K$.]
$$\ \ ^{3}C_{3} + \ ^{4}C_{3} + \ ^{5}C_{3} + \ldots + \ ^{99}C_{3} + \ ^{100}C_{3}{\text{ }K}^{3} = \left( 43n + \frac{101}{4} \right)\ ^{100}C_{3}$$
$$\begin{matrix} & \ \Rightarrow \ ^{4}C_{4} + \ ^{4}C_{3} + \ ^{5}C_{3} + \ldots + \ ^{99}C_{3} + \ ^{100}C_{3}{\text{ }K}^{3} = \left( 43n + \frac{101}{4} \right)^{100}C_{3} \\ & \ = \ ^{100}C_{4} + \ ^{100}C_{3}{\text{ }K}^{3} = \left( 43n + \frac{101}{4} \right)\ ^{100}C_{3} \\ & \ \frac{\ ^{100}C_{4}}{\ ^{100}C_{3}} + K^{3} = 43n + \frac{101}{4} \\ & \ \frac{97}{4} + K^{3} = 43n + \frac{101}{4} \\ & {\text{ }K}^{3} = 43n + 1 \\ & n = 5 \\ & {\text{ }K}^{3} = 216 \\ & \text{ }K = 6 = P \\ & P + n = 11 \end{matrix}$$
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