Finding the nth term of a sequence

[chwala]

Find the nth term of the sequence; $2, 5, 10, 17,26......$ this is how i did it:

2, 5, 10, 17, 26 can form a difference of 3,5, 7,9...
whose nth term is 1 + 2n. now how can apply this to the original sequence to get the nth term?

 

[haruspex]

Find the nth term of the sequence; $2, 5, 10, 17,26......$ this is how i did it:

2, 5, 10, 17, 26 can form a difference of 3,5, 7,9...
whose nth term is 1 + 2n. now how can apply this to the original sequence to get the nth term?

Starting with the 2, you add numbers of the form 2r+1. Do you know how to write a summation expression for a sum of numbers of that form?

 

[chwala]

Starting with the 2, you add numbers of the form 2r+1. Do you know how to write a summation expression for a sum of numbers of that form?

thanks i understand the 2r+1 which is the 1+2n which i indicated, my interest here is how to get the nth term yes sum $Sn= n/2(2a+(n-1)d)$ how do i use this?

 

[haruspex]

thanks i understand the 2r+1 which is the 1+2n which i indicated, my interest here is how to get the nth term yes sum $Sn= n/2(2a+(n-1)d)$ how do i use this?

Find the values of a and d which generate the given sequence.

 

[chwala]

Find the values of a and d which generate the given sequence.

the nth term is $n^2+1$ now my problem is i know $a=2$ what about d? i have tried using $d= 2n+1$ and i am getting $Sn= n/2(2n^2-2+3)$
now how do i move from here to the required $n^2+1$?

 

[haruspex]

i have tried using d=2n+1

No, a and d are constants.
Write the first two terms in the sequence in terms of a and d. That gives you two equations, two unknowns.

 

[chwala]

No, a and d are constants.
Write the first two terms in the sequence in terms of a and d. That gives you two equations, two unknowns.

please assist me i still don't understand.....

 

[haruspex]

please assist me i still don't understand.....

You have an expression Sn=n/2(2a+(n−1)d) which you believe will generate the sequence given the right values of a and d. In particular, it will need to give the right values for S1 and S2. You know what S1 and S2 are, so that gives you two equations.

 

[chwala]

You have an expression Sn=n/2(2a+(n−1)d) which you believe will generate the sequence given the right values of a and d. In particular, it will need to give the right values for S1 and S2. You know what S1 and S2 are, so that gives you two equations.

for S1=n/2(2a+(n-1)d gives
2= 1/2(2.a+(1-1)d) gives 4=2a and S2=2+5=7, 7=2/2(2a+(2-1)d gives 7=2a+d sir, i still don't understand just make it clear for me i am conversant with the general knowledge on arithmetic and geometric...just give the needed equations where a=2 and d=3 how do you move from here to n^2+1?

 

[Ray Vickson]

for S1=n/2(2a+(n-1)d gives
2= 1/2(2.a+(1-1)d) gives 4=2a and S2=2+5=7, 7=2/2(2a+(2-1)d gives 7=2a+d sir, i still don't understand just make it clear for me i am conversant with the general knowledge on arithmetic and geometric...just give the needed equations where a=2 and d=3 how do you move from here to n^2+1?

You were asked to give a formula for the nth term of the sequence; you have done that, so you have answered the question in full. Why do you think you need to "move from here to n^2+1"?

 

[haruspex]

a=2 and d=3

So use those to replace a and d in your expression n/2(2a+(n−1)d).

 

[chwala]

So use those to replace a and d in your expression n/2(2a+(n−1)d).

Thanks i got it......$(1+1)+3+5+7+9+.$...
= $1+(1+3+5+7+9+....)$ forms the series $1+ n^2$

 

[chwala]

This could also be solved by; system of equations. i.e

We have the sequence;
$[ 2,5,10,17,26,...)$
$[3,5,7,9,...)$
$[2,2,2,2,...)$

Since we went up to what i call second difference then the nth term will be quadratic therefore we shall be solving;

$2a=2, a=1$

$3a+b=3 ⇒b=0$

$a+b+c=2 ⇒ 2+0+c=2$

Therefore $c=1$

thus;

$U_n= n^2+1$