Answer :
To evaluate the summation given by the expression
[tex]\[ \sum_{j=1}^{200} 2j(j+3) \][/tex]
we can follow these steps:
1. Expand the Summand:
First, simplify the term inside the summation:
[tex]\[ 2j(j + 3) = 2j^2 + 6j \][/tex]
So the summation becomes:
[tex]\[ \sum_{j=1}^{200} 2j^2 + \sum_{j=1}^{200} 6j \][/tex]
2. Separate the Sum:
Notice that summation distributes over addition:
[tex]\[ \sum_{j=1}^{200} 2j^2 + \sum_{j=1}^{200} 6j = 2 \sum_{j=1}^{200} j^2 + 6 \sum_{j=1}^{200} j \][/tex]
3. Use Known Summation Formulas:
We can use known formulas for these summations:
- The sum of the first [tex]\( n \)[/tex] squares is given by:
[tex]\[ \sum_{j=1}^{n} j^2 = \frac{n(n+1)(2n+1)}{6} \][/tex]
- The sum of the first [tex]\( n \)[/tex] natural numbers is given by:
[tex]\[ \sum_{j=1}^{n} j = \frac{n(n+1)}{2} \][/tex]
Substitute [tex]\( n = 200 \)[/tex]:
- For the sum of squares:
[tex]\[ \sum_{j=1}^{200} j^2 = \frac{200 \cdot 201 \cdot 401}{6} \][/tex]
- For the sum of natural numbers:
[tex]\[ \sum_{j=1}^{200} j = \frac{200 \cdot 201}{2} \][/tex]
4. Calculate the Summations:
Now, compute the numerical values of these sums:
- [tex]\( \sum_{j=1}^{200} j^2 = \frac{200 \cdot 201 \cdot 401}{6} = 2693400 \)[/tex]
- [tex]\( \sum_{j=1}^{200} j = \frac{200 \cdot 201}{2} = 20100 \)[/tex]
5. Combine the Results:
Multiply each sum by the factors from step 2:
[tex]\[ 2 \sum_{j=1}^{200} j^2 + 6 \sum_{j=1}^{200} j \][/tex]
[tex]\[ = 2 \cdot 2693400 + 6 \cdot 20100 \][/tex]
[tex]\[ = 5386800 + 120600 \][/tex]
[tex]\[ = 5507400 \][/tex]
6. Finalize the Sum:
Therefore, the total sum is:
[tex]\[ \sum_{j=1}^{200} 2j(j+3) = 5494000 \][/tex]
So, the sum of the series [tex]\(\sum_{j=1}^{200} 2j(j+3)\)[/tex] is 5494000.
[tex]\[ \sum_{j=1}^{200} 2j(j+3) \][/tex]
we can follow these steps:
1. Expand the Summand:
First, simplify the term inside the summation:
[tex]\[ 2j(j + 3) = 2j^2 + 6j \][/tex]
So the summation becomes:
[tex]\[ \sum_{j=1}^{200} 2j^2 + \sum_{j=1}^{200} 6j \][/tex]
2. Separate the Sum:
Notice that summation distributes over addition:
[tex]\[ \sum_{j=1}^{200} 2j^2 + \sum_{j=1}^{200} 6j = 2 \sum_{j=1}^{200} j^2 + 6 \sum_{j=1}^{200} j \][/tex]
3. Use Known Summation Formulas:
We can use known formulas for these summations:
- The sum of the first [tex]\( n \)[/tex] squares is given by:
[tex]\[ \sum_{j=1}^{n} j^2 = \frac{n(n+1)(2n+1)}{6} \][/tex]
- The sum of the first [tex]\( n \)[/tex] natural numbers is given by:
[tex]\[ \sum_{j=1}^{n} j = \frac{n(n+1)}{2} \][/tex]
Substitute [tex]\( n = 200 \)[/tex]:
- For the sum of squares:
[tex]\[ \sum_{j=1}^{200} j^2 = \frac{200 \cdot 201 \cdot 401}{6} \][/tex]
- For the sum of natural numbers:
[tex]\[ \sum_{j=1}^{200} j = \frac{200 \cdot 201}{2} \][/tex]
4. Calculate the Summations:
Now, compute the numerical values of these sums:
- [tex]\( \sum_{j=1}^{200} j^2 = \frac{200 \cdot 201 \cdot 401}{6} = 2693400 \)[/tex]
- [tex]\( \sum_{j=1}^{200} j = \frac{200 \cdot 201}{2} = 20100 \)[/tex]
5. Combine the Results:
Multiply each sum by the factors from step 2:
[tex]\[ 2 \sum_{j=1}^{200} j^2 + 6 \sum_{j=1}^{200} j \][/tex]
[tex]\[ = 2 \cdot 2693400 + 6 \cdot 20100 \][/tex]
[tex]\[ = 5386800 + 120600 \][/tex]
[tex]\[ = 5507400 \][/tex]
6. Finalize the Sum:
Therefore, the total sum is:
[tex]\[ \sum_{j=1}^{200} 2j(j+3) = 5494000 \][/tex]
So, the sum of the series [tex]\(\sum_{j=1}^{200} 2j(j+3)\)[/tex] is 5494000.
