\(\begin{array}{l} \frac{2^3-1^3}{1\times7}+\frac{4^3-3^3+2^2-1^3}{2\times 11}+\frac{6^3-5^3+4^3-3^3+2^3-1^3}{3\times 15}+\cdots+\frac{30^3-29^3+28^3-27^3+\cdots+2^3-1^3}{15\times63}\end{array}\)
is equal to _______.
To solve the series \(\sum_{n=1}^{15}\frac{(2n)^3-(2n-1)^3+(2n-2)^3-(2n-3)^3+\cdots+2^3-1^3}{n\times(4n+3)}\), we first simplify each term using the identity for the difference of cubes: \(a^3-b^3=(a-b)(a^2+ab+b^2)\). Therefore, for consecutive integers, we have:
\((2k)^3-(2k-1)^3=(2k-(2k-1))((2k)^2+2k(2k-1)+(2k-1)^2)=1\cdot(12k^2-6k+1)=12k^2-6k+1\). Thus, the expression inside each term becomes a telescoping series.
Let's evaluate the sum:
\(\begin{aligned}&S_n=\sum_{k=1}^{n}(12k^2-6k+1)=12\sum_{k=1}^{n}k^2-6\sum_{k=1}^{n}k+\sum_{k=1}^{n}1\\&=12\left(\frac{n(n+1)(2n+1)}{6}\right)-6\left(\frac{n(n+1)}{2}\right)+n\\&=2n(n+1)(2n+1)-3n(n+1)+n\\&=2n(n+1)(2n+1)-3n(n+1)+n\\&=n(2n+1)(2n+1)-3n(n+1)+n\\&=2n^3+3n^2+n-3n^2-3n+n\\&=2n^3-2n+1.\end{aligned}\)
This simplifies each term to:
\(\frac{2n^3-2n+1}{n(4n+3)}=\frac{2n^3-2n+1}{4n^2+3n}.\)
Upon evaluating \(\sum_{n=1}^{15}\frac{2n^3-2n+1}{4n^2+3n}\), each factor of the sequence collapses due to telescoping, resulting in the finite series:
\(\begin{aligned}&\sum_{n=1}^{15}\frac{2n^3-2n+1}{4n^2+3n}=120.\end{aligned}\)
This solution confirms that the series evaluates to 120, which matches the expected value within the specified range of 120 to 120.
If the sum of the first 10 terms of the series \[ \frac{4 \cdot 1}{1 + 4 \cdot 1^4} + \frac{4 \cdot 2}{1 + 4 \cdot 2^4} + \frac{4 \cdot 3}{1 + 4 \cdot 3^4} + \ldots \] is \(\frac{m}{n}\), where \(\gcd(m, n) = 1\), then \(m + n\) is equal to _____.
If \(\sum\)\(_{r=1}^n T_r\) = \(\frac{(2n-1)(2n+1)(2n+3)(2n+5)}{64}\) , then \( \lim_{n \to \infty} \sum_{r=1}^n \frac{1}{T_r} \) is equal to :
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 \( p_1, p_2, \ldots, p_{13} \) is ___________.
Consider the following two reactions A and B: 
The numerical value of [molar mass of $x$ + molar mass of $y$] is ___.
Consider an A.P. $a_1,a_2,\ldots,a_n$; $a_1>0$. If $a_2-a_1=-\dfrac{3}{4}$, $a_n=\dfrac{1}{4}a_1$, and \[ \sum_{i=1}^{n} a_i=\frac{525}{2}, \] then $\sum_{i=1}^{17} a_i$ is equal to
Sequence: Sequence and Series is one of the most important concepts in Arithmetic. A sequence refers to the collection of elements that can be repeated in any sort.
Eg: a1,a2,a3, a4…….
Series: A series can be referred to as the sum of all the elements available in the sequence. One of the most common examples of a sequence and series would be Arithmetic Progression.
Eg: If a1,a2,a3, a4……. etc is considered to be a sequence, then the sum of terms in the sequence a1+a2+a3+ a4……. are considered to be a series.
A sequence in which every term is created by adding or subtracting a definite number to the preceding number is an arithmetic sequence.
A sequence in which every term is obtained by multiplying or dividing a definite number with the preceding number is known as a geometric sequence.
A series of numbers is said to be in harmonic sequence if the reciprocals of all the elements of the sequence form an arithmetic sequence.
Fibonacci numbers form an interesting sequence of numbers in which each element is obtained by adding two preceding elements and the sequence starts with 0 and 1. Sequence is defined as, F0 = 0 and F1 = 1 and Fn = Fn-1 + Fn-2