Asked by Mohamed
                Express following using sigma notation
3+6+11+18+27
            
        3+6+11+18+27
Answers
                    Answered by
            mathhelper
            
    we need a general term for the sequence
Since the consecutive differences between terms is arithmetic,
the sequence itself is quadratic
don't know how good you are at recognizing patterns, but
2+6+11+18+27+..
= 1^2+2 + 2^2+2 + 3^2+2, 4^2+2, 3^2+2, .... , n^2 + 2 , for n = 1,2,3, ....
If you did not see that:
 
let y = ax^2 + bx + c
x = 1, 3 = a + b + c
x = 2, 6 = 4a + 2b + c
x = 3, 11 = 9a + 3b + c
2nd - 3rd: -----> 3a + b = 3
3rd - 2nd: -----> 5a + b = 5
subtract those last two:
2a = 2
a = 1
sub into 3a + b = 3,
3 + b = 3 ----> b = 0
finally into a+b+c = 3
1 + 0 + c = 3 ----> 2
general term is x^2 + 2
3+6+11+18+27 = ∑ (n^2 + 2) for n = 1 to 5
    
Since the consecutive differences between terms is arithmetic,
the sequence itself is quadratic
don't know how good you are at recognizing patterns, but
2+6+11+18+27+..
= 1^2+2 + 2^2+2 + 3^2+2, 4^2+2, 3^2+2, .... , n^2 + 2 , for n = 1,2,3, ....
If you did not see that:
let y = ax^2 + bx + c
x = 1, 3 = a + b + c
x = 2, 6 = 4a + 2b + c
x = 3, 11 = 9a + 3b + c
2nd - 3rd: -----> 3a + b = 3
3rd - 2nd: -----> 5a + b = 5
subtract those last two:
2a = 2
a = 1
sub into 3a + b = 3,
3 + b = 3 ----> b = 0
finally into a+b+c = 3
1 + 0 + c = 3 ----> 2
general term is x^2 + 2
3+6+11+18+27 = ∑ (n^2 + 2) for n = 1 to 5
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