General Term Analysis
The series is given by \( \frac{1}{1+1^{4} \times 4} + \frac{2}{1+2^{4} \times 4} + \frac{3}{1+3^{4} \times 4} + \dots \). The general term, \( T_k \), is:
\( T_k = \frac{k}{1+4k^4} \)
Factor the denominator using the identity \( 1+4x^4 = (1+2x^2+2x)(1+2x^2-2x) \). Substituting \( x=k \):
\( 1+4k^4 = (2k^2 + 2k + 1)(2k^2 - 2k + 1) \)
We express \( T_k \) using the difference of two fractions. Note that \( (2k^2 + 2k + 1) - (2k^2 - 2k + 1) = 4k \).
\( T_k = \frac{1}{4} \times \frac{4k}{(2k^2 - 2k + 1)(2k^2 + 2k + 1)} \) \( T_k = \frac{1}{4} \left( \frac{1}{2k^2 - 2k + 1} - \frac{1}{2k^2 + 2k + 1} \right) \)
Telescoping Series Summation
Define \( f(k) = 2k^2 - 2k + 1 \). Then \( f(k+1) = 2(k+1)^2 - 2(k+1) + 1 = 2(k^2+2k+1) - 2k - 2 + 1 = 2k^2 + 2k + 1 \).
The general term simplifies to a telescoping form: \( T_k = \frac{1}{4} \left( \frac{1}{f(k)} - \frac{1}{f(k+1)} \right) \)
The sum of the first 10 terms, \( S_{10} \), is: \( S_{10} = \sum_{k=1}^{10} T_k = \frac{1}{4} \sum_{k=1}^{10} \left( \frac{1}{f(k)} - \frac{1}{f(k+1)} \right) \) This sum telescopes to: \( S_{10} = \frac{1}{4} \left( \frac{1}{f(1)} - \frac{1}{f(11)} \right) \)
Calculating the Sum \( \frac{m}{n} \)
First, compute \( f(1) \) and \( f(11) \): \( f(1) = 2(1)^2 - 2(1) + 1 = 1 \) \( f(11) = 2(11)^2 - 2(11) + 1 = 2(121) - 22 + 1 = 242 - 21 = 221 \)
Now, calculate \( S_{10} \): \( S_{10} = \frac{1}{4} \left( \frac{1}{1} - \frac{1}{221} \right) = \frac{1}{4} \left( \frac{221 - 1}{221} \right) = \frac{1}{4} \times \frac{220}{221} = \frac{55}{221} \)
So, \( \frac{m}{n} = \frac{55}{221} \).
Finding \( m+n \)
We have \( m = 55 \) and \( n = 221 \). Check for the greatest common divisor (GCD): \( m = 55 = 5 \times 11 \) \( n = 221 = 13 \times 17 \)
The prime factors of \( m \) and \( n \) are distinct, so \( \operatorname{gcd}(55, 221) = 1 \).
The value required is \( m+n \): \( m+n = 55 + 221 = 276 \)
Similar Questions
- Let A be a $3 \times 3$ matrix such that $A + A^T = O$. If $A\begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} = \begin{bmatrix} 3 \\ 3 \\ 2 \end{bmatrix}$, $A^2\begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} = \begin{bmatrix} -3 \\ 19 \\ -24 \end{bmatrix}$ and $\det(adj(2 \ adj(A + I))) = (2)^\alpha \cdot (3)^\beta \cdot (11)^\gamma$, $\alpha, \beta, \gamma$ are non-negative integers, then $\alpha + \beta + \gamma$ is equal to _________
- Let $\alpha = \frac{-1 + i\sqrt{3}}{2}$ and $\beta = \frac{-1 - i\sqrt{3}}{2}$, $i = \sqrt{-1}$. If $(7 - 7\alpha + 9\beta)^{20} + (9 + 7\alpha - 7\beta)^{20} + (-7 + 9\alpha + 7\beta)^{20} + (14 + 7\alpha + 7\beta)^{20} = m^{10}$, then $m$ is _________
- Let ABC be a triangle. Consider four points $p_1, p_2, p_3, p_4$ on the side AB, five points $p_5, p_6, p_7, p_8, p_9$ on the side BC, and four points $p_{10}, p_{11}, p_{12}, p_{13}$ on the side AC. None of these points is a vertex of the triangle ABC. Then the total number of pentagons, that can be formed by taking all the vertices from the points $p_1, p_2, ..., p_{13}$, is _________
- If $X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}$ is a solution of the system of equations $AX = B$, where $\text{adj } A = \begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3 \end{bmatrix}$ and $B = \begin{bmatrix} 4 \\ 0 \\ 2 \end{bmatrix}$, then $|x + y + z|$ is equal to :
- Let $C_r$ denote the coefficient of $x^r$ in the binomial expansion of $(1 + x)^n$, $n \in \mathbb{N}, 0 \leq r \leq n$. If $P_n = C_0 - C_1 + \frac{2^2}{3} C_2 - \frac{2^3}{4} C_3 + \dots + \frac{(-2)^n}{n+1} C_n$, then the value of $\sum_{n=1}^{25} \frac{1}{P_{2n}}$ equals.
- The number of elements in the relation $R = \{(x, y) : 4x^2 + y^2 < 52, x, y \in \mathbb{Z}\}$ is
- Let $S = \{z \in \mathbb{C} : 4z^2 + \bar{z} = 0\}$. Then $\sum_{z \in S} |z|^2$ is equal to :
Let f and g be functions satisfying $f(x+y) = f(x)f(y), f(1) = 7$ and $g(x+y) = g(xy), g(1) = 1$, for all $x, y \in \mathbb{N}$. If $\sum_{x=1}^{n} \left(\frac{f(x)}{g(x)}\right) = 19607$, then n is equal to :
Let S be the set of the first 11 natural numbers. Then the number of elements in $A = \{B \subseteq S : n(B) \geq 2$ and the product of all elements of B is even is ________.
- Let $S = \frac{1}{25!} + \frac{1}{3!23!} + \frac{1}{5!21!} + \dots$ up to 13 terms. If $13S = \frac{2^k}{n!}, k \in \mathbb{N}$, then $n + k$ is equal to
Important Questions from Algebra
- Let A be a $3 \times 3$ matrix such that $A + A^T = O$. If $A\begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} = \begin{bmatrix} 3 \\ 3 \\ 2 \end{bmatrix}$, $A^2\begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} = \begin{bmatrix} -3 \\ 19 \\ -24 \end{bmatrix}$ and $\det(adj(2 \ adj(A + I))) = (2)^\alpha \cdot (3)^\beta \cdot (11)^\gamma$, $\alpha, \beta, \gamma$ are non-negative integers, then $\alpha + \beta + \gamma$ is equal to _________
- Let $\alpha = \frac{-1 + i\sqrt{3}}{2}$ and $\beta = \frac{-1 - i\sqrt{3}}{2}$, $i = \sqrt{-1}$. If $(7 - 7\alpha + 9\beta)^{20} + (9 + 7\alpha - 7\beta)^{20} + (-7 + 9\alpha + 7\beta)^{20} + (14 + 7\alpha + 7\beta)^{20} = m^{10}$, then $m$ is _________
- Let ABC be a triangle. Consider four points $p_1, p_2, p_3, p_4$ on the side AB, five points $p_5, p_6, p_7, p_8, p_9$ on the side BC, and four points $p_{10}, p_{11}, p_{12}, p_{13}$ on the side AC. None of these points is a vertex of the triangle ABC. Then the total number of pentagons, that can be formed by taking all the vertices from the points $p_1, p_2, ..., p_{13}$, is _________
- If $X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}$ is a solution of the system of equations $AX = B$, where $\text{adj } A = \begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3 \end{bmatrix}$ and $B = \begin{bmatrix} 4 \\ 0 \\ 2 \end{bmatrix}$, then $|x + y + z|$ is equal to :
- Let $C_r$ denote the coefficient of $x^r$ in the binomial expansion of $(1 + x)^n$, $n \in \mathbb{N}, 0 \leq r \leq n$. If $P_n = C_0 - C_1 + \frac{2^2}{3} C_2 - \frac{2^3}{4} C_3 + \dots + \frac{(-2)^n}{n+1} C_n$, then the value of $\sum_{n=1}^{25} \frac{1}{P_{2n}}$ equals.
