01 Summation Notation

01 Summation Notation#

Enter closed-form expressions, without summation notation, in terms of the variables given in each question.

Question 01#

\[ \sum_{i=0}^{N} 1 \]

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#

\[ \sum _{k=1}^{K} \sum _{t=1}^{T} 1 \]

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:

\[ \sum _{k=1}^{K} \sum _{t=1}^{T} 1 = K \times T \]

Question 03#

\[ \sum_{k=1}^{K} \sum_{t=1}^{T} 0.5^k \]

Explnation This expression consists of two summations:

  1. The outer summation, indexed by ( k ), running from 1 to ( K ).

  2. 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:

  1. 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 ).

    \[ \sum_{t=1}^{T} 0.5^k = T \cdot 0.5^k \]
  2. 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:

    \[ T \cdot \sum_{k=1}^{K} 0.5^k \]
  3. 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 ):

    \[ \sum_{k=1}^{K} 0.5^k = 0.5 \frac{1 - 0.5^K}{1 - 0.5} = 0.5 \frac{1 - 0.5^K}{0.5} = 1 - 0.5^K \]
  4. Combining Results: Substitute this result back into our expression:

    \[ T \cdot (1 - 0.5^K) \]

So, the double summation simplifies to:

\[ \sum_{k=1}^{K} \sum_{t=1}^{T} 0.5^k = T \cdot (1 - 0.5^K) \]

This is the simplified form of the given double summation.

Question 04#

\[ \sum_{k=1}^{\infty} \sum_{t=1}^{T} 0.5^k \]

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:

  1. 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 ):

    \[ \sum_{t=1}^{T} 0.5^k = T \cdot 0.5^k \]
  2. 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:

    \[ T \cdot \sum_{k=1}^{\infty} 0.5^k \]
  3. 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 ):

    \[ \sum_{k=1}^{\infty} 0.5^k = \frac{0.5}{1 - 0.5} = \frac{0.5}{0.5} = 1 \]
  4. Combining Results: Substitute this result back into our expression:

    \[ T \cdot 1 = T \]

So, the double summation simplifies to:

\[ \sum_{k=1}^{\infty} \sum_{t=1}^{T} 0.5^k = T \]

This is the simplified form of the given double summation.