Let the smallest value of kN, for which the coefficient of in , , is for some , be p. Then the value of p + n is: [2026]

Question

Let the smallest value of k $\in$ N, for which the coefficient of $x^{3}$ in ${(1 + x)}^{3} + {(1 + x)}^{4} + {(1 + x)}^{5} + . . . + {(1 + x)}^{99} + {(1 + k x)}^{100}$ , $x \neq 0$ , is $(43 n + \frac{101}{4}) ({}^{100}C_{3})$ for some $n \in N$ , be p. Then the value of p + n is: [2026]

Smart Achievers · Accepted Answer

(2)

Coeff. of $x^{3} = {}^{3}C_{3} + {}^{4}C_{3} + {}^{5}C_{3} + . . . + {}^{99}C_{3} + {}^{100}C_{3} \cdot k^{3}$

$= {}^{4}C_{4} + {}^{4}C_{3} + {}^{5}C_{3} + . . . + {}^{99}C_{3} + {}^{100}C_{3} \cdot k^{3}$

Using ${}^{n}C_{r} + {}^{n}C_{r - 1} = {}^{n + 1}C_{r}$

Coeff. of $x^{3} = {}^{100}C_{4} + {}^{100}C_{3} \cdot k^{3}$

$= \frac{100!}{4! \times 96!} + \frac{100!}{97! \times 3!} k^{3}$

$= \frac{100!}{97! \times 3!} {\frac{97}{4} + k^{3}} = (43 n + \frac{101}{4}) {}^{100}C_{3}$ (Given)

$\Rightarrow k^{3} - 43 n = 1$

If n = 1, then $k^{3}$ = 44

If n = 2, then $k^{3}$ = 87

If n = 5, then $k^{3}$ = 216

$\Rightarrow$ k = 6 (smallest k)

$∴$ p = 6 and n = 5

Thus, p + n = 11

Solution

Q.
Let the smallest value of k $\in$ N, for which the coefficient of $x^{3}$ in ${(1 + x)}^{3} + {(1 + x)}^{4} + {(1 + x)}^{5} + . . . + {(1 + x)}^{99} + {(1 + k x)}^{100}$ , $x \neq 0$ , is $(43 n + \frac{101}{4}) ({}^{100}C_{3})$ for some $n \in N$ , be p. Then the value of p + n is: [2026]

1 10

2 11

3 12

4 13

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