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 = {-3, -2, -1, 0, 1, 2, 3}. Let R be a relation on A defined by xRy if and only if $0\le x^2+2y\le 4$. Let $l$ be the number of elements in R and $m$ be the minimum number of elements required to be added in R to make it a reflexive relation. Then $l + m$ is equal to
Let $\alpha$ and $\beta$ be the roots of $x^2 + \sqrt{3}x-16=0$, and $\gamma$ and $\delta$ be the roots of $x^2 + 3x - 1 = 0$. If $P_n = \alpha^n + \beta^n$ and $Q_n = \gamma^n + \delta^n$, then $\frac{P_{25} + \sqrt{3}P_{24}}{2P_{23}} + \frac{Q_{25}-Q_{23}}{Q_{24}}$ is equal to
Let the domain of the function $f (x) = \log_2 \log_4 \log_6(3 + 4x -x^2)$ be $(a, b)$.
If $\int_{b-a}^{b+a}[x^2] dx=p-\sqrt{q}-\sqrt{r}$, $p,q,r\in N$, $gcd(p,q,r) = 1$, where $[.]$ is the greatest integer function, then $p + q + r$ is equal to
Let A be a matrix of order $3 \times 3$ and $|A| = 5$. If $|2\text{adj} (3A \text{adj} (2A))| = 2^\alpha \cdot 3^\beta \cdot 5^\gamma$, $\alpha, \beta, \gamma \in N$, then $\alpha + \beta + \gamma$ is equal to
Let $a_1, a_2, a_3,....$ be a G.P. of increasing positive numbers. If $a_3a_5 = 729$ and $a_2 + a_4 = \frac{111}{4}$, then $24 (a_1 + a_2 + a_3)$ is equal to
All five letter words are made using all the letters A, B, C, D, E and arranged as in an English dictionary with serial numbers. Let the word at serial number $n$ be denoted by $W_n$. Let the probability $P(W_n)$ of choosing the word $W_n$ satisfy $P(W_n) = 2P(W_{n-1})$, $n > 1$.
If $P(CDBEA) = \frac{2^\alpha}{2^\beta-1}$, $\alpha, \beta\in N$, then $\alpha + \beta$ is equal to :
- The largest $n \in N$ such that $3^n$ divides $50!$ is :
- The number of sequences of ten terms, whose terms are either 0 or 1 or 2, that contain exactly five 1s and exactly three 2s, is equal to :
- Let A be the set of all functions $f: Z \to Z$ and R be a relation on A such that $R = \{(f, g) : f(0) = g(1) \text{ and } f(1) = g(0)\}$. Then R is :
Let $a \in \mathbf{R}$ and $A$ be a matrix of order $3 \times 3$ such that $\det (A) = -4$ and $A + I = \begin{bmatrix} 1 & a & 1 \\ 2 & 1 & 0 \\ a & 1 & 2 \end{bmatrix}$, where $I$ is the identity matrix of order $3 \times 3$. If $\det ((a+1)\text{adj}((a-1)A))$ is $2^m 3^n$, $m, n \in \{0, 1, 2, \dots, 20\}$, then $m+n$ is equal to :
Important Questions from Algebra
Let A = {-3, -2, -1, 0, 1, 2, 3}. Let R be a relation on A defined by xRy if and only if $0\le x^2+2y\le 4$. Let $l$ be the number of elements in R and $m$ be the minimum number of elements required to be added in R to make it a reflexive relation. Then $l + m$ is equal to
Let $\alpha$ and $\beta$ be the roots of $x^2 + \sqrt{3}x-16=0$, and $\gamma$ and $\delta$ be the roots of $x^2 + 3x - 1 = 0$. If $P_n = \alpha^n + \beta^n$ and $Q_n = \gamma^n + \delta^n$, then $\frac{P_{25} + \sqrt{3}P_{24}}{2P_{23}} + \frac{Q_{25}-Q_{23}}{Q_{24}}$ is equal to
Let the domain of the function $f (x) = \log_2 \log_4 \log_6(3 + 4x -x^2)$ be $(a, b)$.
If $\int_{b-a}^{b+a}[x^2] dx=p-\sqrt{q}-\sqrt{r}$, $p,q,r\in N$, $gcd(p,q,r) = 1$, where $[.]$ is the greatest integer function, then $p + q + r$ is equal to
Let A be a matrix of order $3 \times 3$ and $|A| = 5$. If $|2\text{adj} (3A \text{adj} (2A))| = 2^\alpha \cdot 3^\beta \cdot 5^\gamma$, $\alpha, \beta, \gamma \in N$, then $\alpha + \beta + \gamma$ is equal to
Let $a_1, a_2, a_3,....$ be a G.P. of increasing positive numbers. If $a_3a_5 = 729$ and $a_2 + a_4 = \frac{111}{4}$, then $24 (a_1 + a_2 + a_3)$ is equal to
