Question:
Which summation formula represents the series below? 1 + 2 + 6 + 24
(a) [tex]\sum_{n=2}^{5}(n-1) ![/tex]
(b) [tex]\sum_{n=0}^{3} n ![/tex]
(c) [tex]\sum_{n=1}^{4}(n+1) ![/tex]
(d) [tex]\sum_{n=2}^{5} n ![/tex]
Answer:
Option a: [tex]\sum_{n=2}^{5}(n-1) ![/tex] is the correct answer.
Explanation:
Option a: [tex]\sum_{n=2}^{5}(n-1) ![/tex]
By substituting the values of n and expanding the summation, we have,
[tex](2-1) !+(3-1) !+(4-1) !+(5-1) ![/tex]
Subtracting, we have,
[tex]1 !+2!+3 !+4 ![/tex]
Expanding the factorial,
[tex]1+(2*1)+(3*2*1)+(4*3*2*1)[/tex]
Simplifying, we get,
[tex]1+2+6+24[/tex]
Thus, the summation [tex]\sum_{n=2}^{5}(n-1) ![/tex] represents the series [tex]1+2+6+24[/tex]
Hence, Option a is the correct answer.
Option b: [tex]\sum_{n=0}^{3} n ![/tex]
By substituting the values of n and expanding the summation, we have,
[tex]0!+1!+2!+3![/tex]
Expanding the factorial,
[tex]0+1+(2*1)+(3*2*1)[/tex]
Simplifying, we get,
[tex]0+1+2+6[/tex]
Thus, the summation [tex]\sum_{n=0}^{3} n ![/tex] does not represents the series [tex]1+2+6+24[/tex]
Hence, Option b is not the correct answer.
Option c: [tex]\sum_{n=1}^{4}(n+1) ![/tex]
By substituting the values of n and expanding the summation, we have,
[tex](1+1) !+(2+1) !+(3+1) !+(4+1) ![/tex]
Adding, we have,
[tex]2!+3!+4!+5![/tex]
Expanding the factorial,
[tex](2*1)+(3*2*1)+(4*3*2*1)+(5*4*3*2*1)[/tex]
Simplifying, we get,
[tex]2+6+24+120[/tex]
Thus, the summation [tex]\sum_{n=1}^{4}(n+1) ![/tex] does not represents the series [tex]1+2+6+24[/tex]
Hence, Option c is not the correct answer.
Option d: [tex]\sum_{n=2}^{5} n ![/tex]
By substituting the values of n and expanding the summation, we have,
[tex]2!+3!+4!+5![/tex]
Expanding the factorial,
[tex](2*1)+(3*2*1)+(4*3*2*1)+(5*4*3*2*1)[/tex]
Simplifying, we get,
[tex]2+6+24+120[/tex]
Thus, the summation [tex]\sum_{n=2}^{5} n ![/tex] does not represents the series [tex]1+2+6+24[/tex]
Hence, Option d is not the correct answer.
Hence, the correct answer is Option a: [tex]\sum_{n=2}^{5}(n-1) ![/tex]
