what is the nth term of given sequence?

Question

i want to find nth term and sum of n terms of 6+13+24+39+...........

Answer 1

Series: 6, 13, 24, 39, 58

1st diff: 7, 11, 15, 19

2nd diff: 4, 4, 4

An extra term has been added to the series to illustrate what appears to be the pattern.

The second difference is constant and this suggest a quadratic function for the nth term, a_n.

Let the quadratic function be f(n)=an²+bn+c where a, b and c are constant coefficients.

f(1)=6=a+b+c

f(2)=13=4a+2b+c

f(3)=24=9a+3b+c

Now we have 3 equations and 3 unknowns so we can calculate a, b and c.

f(2)-f(1)=7=3a+b;

f(3)-f(2)=11=5a+b.

Therefore 11-7=4=2a, so a=2, b=1, c=3. f(n)=a_n=2n²+n+3.

f(4)=2×16+4+3=39 which matches expectation from the series.

Also, f(5)=2×25+5+3=58.

To find the sum of the series, S_n=2∑n²+∑n+3n.

Sum of the natural numbers: n(n+1)/2.

Sum of the squares of natural numbers: 1, 5, 14, 30, 66,...=n(n+1)(2n+1)/6.

S_n=2n(n+1)(2n+1)/6+n(n+1)/2+3n,

S_n=(2n(n+1)(2n+1)+3n(n+1)+18n)/6,

S_n=n(n+1)(4n+2+3+18/(n+1))/6,

S_n=n(n+1)(4n+5+18/(n+1))/6,

S_n=n(4n²+9n+23)/6.

S₁=6, S₂=19, S₃=43, S₄=82; S₅=140 from the formula.

S₂-S₁=a₂=13; S₃-S₂=a₃=24; S₄-S₃=a₄=39; S₅-S₄=a₅=58.

These match the series.