Question

If $1(50)^{49} + 2(51)^1(50)^{48} + 3(51)^2(50)^{47} + \dots + (50)(51)^{49} = k(50)^{49}$ then the number of factor of k of the form 4m + 1, ($m \in W$) is/are

• A. 3
• B. 5
• C. 4
• D. 7

Answer 1

Answer: (B)

5

Answer 2

The given series is $S = \sum_{n=1}^{50} n (51)^{n-1} (50)^{50-n}$. Let $a=50$ and $b=51$. The series is $S = \sum_{n=1}^{50} n b^{n-1} a^{50-n}$. Factoring out $a^{49}$, we get $S = a^{49} \left[ 1 + 2\left(\frac{b}{a}\right) + 3\left(\frac{b}{a}\right)^2 + \dots + 50\left(\frac{b}{a}\right)^{49} \right]$. Let $r = \frac{b}{a} = \frac{51}{50}$. Let $T = 1 + 2r + 3r^2 + \dots + 50r^{49}$. This is an arithmetic-geometric series. Consider the geometric series $G(r) = 1 + r + r^2 + \dots + r^{50} = \frac{r^{51}-1}{r-1}$. Differentiating with respect to $r$, we get $T = G'(r) = \frac{d}{dr}\left(\frac{r^{51}-1}{r-1}\right) = \frac{50r^{51} - 51r^{50} + 1}{(r-1)^2}$. Substituting $r = \frac{51}{50}$, we find $r-1 = \frac{1}{50}$ and $(r-1)^2 = \frac{1}{2500}$. $T = \frac{50(\frac{51}{50})^{51} - 51(\frac{51}{50})^{50} + 1}{(\frac{1}{50})^2} = 2500 \left[ 50 \cdot \frac{51^{51}}{50^{51}} - 51 \cdot \frac{51^{50}}{50^{50}} + 1 \right]$ $T = 2500 \left[ \frac{51^{51}}{50^{50}} - \frac{51^{51}}{50^{50}} + 1 \right] = 2500 [0 + 1] = 2500$. So, $S = a^{49} \cdot T = (50)^{49} \cdot 2500$. Given $S = k(50)^{49}$, we have $k = 2500$. The prime factorization of $k=2500$ is $2^2 \times 5^4$. A factor $d$ of $k$ is of the form $2^a \times 5^b$, where $0 \le a \le 2$ and $0 \le b \le 4$. We need factors $d$ such that $d \equiv 1 \pmod{4}$. $d \equiv (2^a \pmod{4}) \times (5^b \pmod{4}) \pmod{4}$. Since $5 \equiv 1 \pmod{4}$, $5^b \equiv 1^b \equiv 1 \pmod{4}$. So, $d \equiv 2^a \pmod{4}$. For $d \equiv 1 \pmod{4}$, we need $2^a \equiv 1 \pmod{4}$. This is true only when $a=0$. The factors of the form $4m+1$ are of the form $2^0 \times 5^b = 5^b$, where $b \in \{0, 1, 2, 3, 4\}$. There are 5 possible values for $b$ (0, 1, 2, 3, 4), so there are 5 factors of $k$ of the form $4m+1$. These factors are $5^0=1$, $5^1=5$, $5^2=25$, $5^3=125$, and $5^4=625$. All are of the form $4m+1$ for $m \in W$.