Let the smallest value of k N, for which the coefficient of x^3 in (1+x)^3+(1+x)^4+(1+x)^5…

Question

Let the smallest value of $k\in N$, for which the coefficient of $x^3$ in $(1+x)^3+(1+x)^4+(1+x)^5+\cdots+(1+x)^{99}+(1+kx)^{100}$ is $\left(43n+\dfrac{101}{4}\right)\left({}^{100}C_3\right)$ for some $n\in N$, be $p$. Then the value of $p+n$ is:

Accepted Answer

The coefficient of $x^3$ in $(1+x)^3+(1+x)^4+\cdots+(1+x)^{99}$ is $\displaystyle\sum_{r=3}^{99}\binom{r}{3} = \binom{100}{4}$ (hockey-stick identity), and the coefficient of $x^3$ in $(1+kx)^{100}$ is $\binom{100}{3}k^3$.

So total coefficient of $x^3$ is $\binom{100}{4}+\binom{100}{3}k^3$. Setting this equal to $\left(43n+\dfrac{101}{4}\right)\binom{100}{3}$:

$$\frac{\binom{100}{4}}{\binom{100}{3}}+k^3 = 43n+\frac{101}{4}$$

Since $\dfrac{\binom{100}{4}}{\binom{100}{3}}=\dfrac{100-3}{4}=\dfrac{97}{4}$, this gives $k^3+\dfrac{97}{4}=43n+\dfrac{101}{4}$, i.e.

$k^3 = 43n+1$

We need the smallest $k\in N$ ($k\ge1$) such that $k^3-1$ is divisible by 43 for some $n\ge1$. Testing $n=1,2,3,4,5,\dots$: $k^3=44,87,130,173,216,\dots$ Only $n=5$ gives $k^3=216=6^3$, so $k=6$.

Thus $p=6$, $n=5$, and $p+n=11$.

Solutions · 1