what is the nth term in the series 1,7,19,37? please help me...?

Answer 1

The formula could be many things. One possibility is a(n) = 3n^2 - 3n + 1

Answer 2

1,7,19, 37.....

In above series the differences increase by 6 ie.. difference between first & second is 6, second & third is 12...so on....
1 + 6 = 7 (n = 2)
1 + 6 + 2*6 = 19 (n = 3)
1+ 6 + 2*6 + 3*6 =37 (n = 4)
.......
Formulae for nth term is 1 + 6 ( 1+2+3+..+(n-1))

1+2+3+...+n = n*(n+1)/2
1+2+3+...+(n-1) = (n-1)*(n-1+1)/2 = (n-1)*n/2

So the formula becomes

1 + 6(n-1)*n/2 ---> 1 + 3(n-1)*n (i.e 6/2 = 3)

Finally, the answer is

1+ 3(n-1)*n

For example n = 3, 1 + 3*(3-1)*3 = 19
For example n = 4, 1 + 3*(4-1)*4 = 37
.......................
For example n = n, 1 + 3*(n-1)*n = value depends on 'n'

Answer 3

a(n)=1+(6*(n-1))

a(n) is the nth term in the series

Answer 4

x[n] = x[n - 1] + 6(n - 1)

Answer 5

Is it the ninth term in the series? It is going up by 6 each time therefore 6,12,18,24,30,36,42,48,54 and then added them up to the last number:
1,2,19,37,61,91,127,169,217,271 Ans. 271

Answer 6

These are all prime numbers. Only every 4th prime number is in the seres.

http://en.wikipedia.org/wiki/Prime_number