Question

pattern 1,5,11,19 1.how do i find the next two patterns 2.how do i calculate the n^th term of the pattern

Answer 1

form 4 columns
x, y, first difference in y's, 2nd difference in y's

1 1
2 5 4
3 11 6 2
4 19 8 2

since the 2nd differences are constant, we have a quadratic
let the relation be
term(n) = an^2 + bn + c
for (1,1) --- 1 = a + b + c
for (2,5) --- 5 = 4a + 2b + c
for (3,11) --- 11 = 9a + 3b + c

subtract the first two:
3a + b = 4
subtract the 2nd from the 3rd:
5a + b = 6
subtract those last two:
2a = 2
a = 1
sub into 3a+b = 4 ---> b = 1
sub into a+b+c = 1
1 + 1 + c = 1
c = -1

rule: term(n) = n^2 + n - 1


my first 3 values are satisfied by this equation, let's check if the last one works
if x = 4
term(4) = 16 + 4 - 1 = 19

yeahhh!!!!

Answer 2

check the differences: 4,6,8
I suspect that the next terms are

19+10=29 and 29+12 = 41

you plug in 3 pairs of values into a general quadratic y=ax^2+bx+c and you get

a+b+c=1
4a+2b+c=5
9a+3b+c=11
a=1, b=1, c=-1
and get Tn = f(n) = x^2+x-1

Or, you can recall that

1+2+3+...n = n(n+1)/2
You have

1 + 4+6+8+...
= -1 + 2+4+6+...
= -1+2(1+2+3+...)
Tn = -1+2*n(n+1)/2
= -1 + n(n+1)
= n^2+n-1

Answer 3

thanks which term of the pattern
is equal to 2549

Answer 4

come on. You have the formula. Just solve

n^2+n-1 = 2549

Hint: it's not -51