01 Summation Notation#
Enter closed-form expressions, without summation notation, in terms of the variables given in each question.
Question 01#
Explanation To evaluate the summation \(\sum_{i=0}^{N} 1\), we can observe that the term being summed is always equal to 1. Since we are summing the term 1 for each value of \(i\) from 0 to (N), the result will be the total number of terms in the summation, which is (N+1).
Therefore, the simplified form of the given summation is: $\( \sum_{i=0}^{N} 1 = N+1 \)$
Question 02#
Explanation To evaluate the given double summation, we can observe that the term being summed is always equal to 1. Since we are summing the term 1 for each value of (k) from 1 to (K) and for each value of (t) from 1 to (T), the result will be the total number of terms in the summation, which is \(K \times T\).
Therefore, the simplified form of the given double summation is:
Question 03#
Explnation This expression consists of two summations:
The outer summation, indexed by ( k ), running from 1 to ( K ).
The inner summation, indexed by ( t ), running from 1 to ( T ).
The term being summed is ( 0.5^k ), which depends only on the outer index ( k ).
Here’s the step-by-step breakdown:
Inner Summation: The inner summation is over ( t ), but the term ( 0.5^k ) does not depend on ( t ). This means that for each fixed ( k ), the inner summation is just ( T ) times ( 0.5^k ).
\[ \sum_{t=1}^{T} 0.5^k = T \cdot 0.5^k \]Outer Summation: Now, we sum the result of the inner summation over ( k ) from 1 to ( K ).
\[ \sum_{k=1}^{K} T \cdot 0.5^k \]Since ( T ) is a constant with respect to ( k ), it can be factored out:
\[ T \cdot \sum_{k=1}^{K} 0.5^k \]Geometric Series: The summation ( \sum_{k=1}^{K} 0.5^k ) is a geometric series with the first term ( a = 0.5 ) and common ratio ( r = 0.5 ).
The sum of the first ( K ) terms of a geometric series is given by:
\[ \sum_{k=1}^{K} r^k = r \frac{1 - r^K}{1 - r} \]Plugging in ( r = 0.5 ):
\[ \sum_{k=1}^{K} 0.5^k = 0.5 \frac{1 - 0.5^K}{1 - 0.5} = 0.5 \frac{1 - 0.5^K}{0.5} = 1 - 0.5^K \]Combining Results: Substitute this result back into our expression:
\[ T \cdot (1 - 0.5^K) \]
So, the double summation simplifies to:
This is the simplified form of the given double summation.
Question 04#
Explanation This expression is similar to the previous one, with the difference that the outer summation now runs to infinity ((\infty)) instead of a finite upper limit (K). Let’s break it down:
Inner Summation: The inner summation is still over ( t ), but the term ( 0.5^k ) does not depend on ( t ). This means that for each fixed ( k ), the inner summation is just ( T ) times ( 0.5^k ):
\[ \sum_{t=1}^{T} 0.5^k = T \cdot 0.5^k \]Outer Summation: Now, we sum the result of the inner summation over ( k ) from 1 to infinity:
\[ \sum_{k=1}^{\infty} T \cdot 0.5^k \]Since ( T ) is a constant with respect to ( k ), it can be factored out:
\[ T \cdot \sum_{k=1}^{\infty} 0.5^k \]Geometric Series: The summation ( \sum_{k=1}^{\infty} 0.5^k ) is an infinite geometric series with the first term ( a = 0.5 ) and common ratio ( r = 0.5 ). The sum of an infinite geometric series ( \sum_{k=0}^{\infty} ar^k ) is given by:
\[ \sum_{k=1}^{\infty} r^k = \frac{r}{1 - r} \]Plugging in ( r = 0.5 ):
\[ \sum_{k=1}^{\infty} 0.5^k = \frac{0.5}{1 - 0.5} = \frac{0.5}{0.5} = 1 \]Combining Results: Substitute this result back into our expression:
\[ T \cdot 1 = T \]
So, the double summation simplifies to:
This is the simplified form of the given double summation.