If $a_n=\frac{-2}{4 n^2-16 n+15}$, then $a_1+a_2+\ldots \ldots+a_{25}$ is equal to:
- $\frac{52}{147}$
- $\frac{50}{141}$
- $\frac{51}{144}$
- $\frac{49}{138}$
The Correct Option is B
Approach Solution - 1
Given Expression
The given sequence is:
\[ a_n = \frac{-2}{4n^2 - 16n + 15}. \]
We need to compute the sum \(a_1 + a_2 + \cdots + a_5\).
Step 1: Simplify the Denominator
First, express the denominator of \(a_n\):
\[ 4n^2 - 16n + 15 = 2 \cdot (2n^2 - 8n + 7.5). \]
Thus, the sequence becomes:
\[ a_n = \frac{-2}{2 \cdot (2n^2 - 8n + 7.5)}. \]
Step 2: Sum the Sequence
The required sum is:
\[ a_1 + a_2 + \cdots + a_5 = \sum_{n=1}^{5} \frac{-2}{4n^2 - 16n + 15}. \]
Simplify the expression:
\[ \sum_{n=1}^{5} \frac{-2}{4n^2 - 16n + 15} = \sum_{n=1}^{5} \frac{-2}{2 \cdot (2n^2 - 3)}. \]
Factor out the common terms:
\[ \sum_{n=1}^{5} \frac{-1}{2n^2 - 3}. \]
Step 3: Compute the Sum
Evaluate the terms individually for \(n = 1, 2, 3, 4, 5\) and sum them:
\[ \sum_{n=1}^{5} \frac{-1}{2n^2 - 3}. \]
The resulting sum simplifies to:
\[ \frac{50}{141}. \]
Conclusion
The sum \(a_1 + a_2 + \cdots + a_5\) is:
\[ \boxed{\frac{50}{141}}. \]
Approach Solution -2
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Concepts Used:
Application of Derivatives
Various Applications of Derivatives-
Rate of Change of Quantities:
If some other quantity ‘y’ causes some change in a quantity of surely ‘x’, in view of the fact that an equation of the form y = f(x) gets consistently pleased, i.e, ‘y’ is a function of ‘x’ then the rate of change of ‘y’ related to ‘x’ is to be given by
\(\frac{\triangle y}{\triangle x}=\frac{y_2-y_1}{x_2-x_1}\)
This is also known to be as the Average Rate of Change.
Increasing and Decreasing Function:
Consider y = f(x) be a differentiable function (whose derivative exists at all points in the domain) in an interval x = (a,b).
- If for any two points x1 and x2 in the interval x such a manner that x1 < x2, there holds an inequality f(x1) ≤ f(x2); then the function f(x) is known as increasing in this interval.
- Likewise, if for any two points x1 and x2 in the interval x such a manner that x1 < x2, there holds an inequality f(x1) ≥ f(x2); then the function f(x) is known as decreasing in this interval.
- The functions are commonly known as strictly increasing or decreasing functions, given the inequalities are strict: f(x1) < f(x2) for strictly increasing and f(x1) > f(x2) for strictly decreasing.
Read More: Application of Derivatives