Term 1 = 9
= 6 + 3
= 6 + 1 x 3
= 6 + 1 x (1 + 2)

Term 2 = 14
= 6 + 8
= 6 + 2 x 4
= 6 + 2 x (2 + 2)


Term 3 = 21
= 6 + 15
= 6 + 3 x 5
= 6 + 3 x (3 + 2)


Term 4 = 30
= 6 + 24
= 6 + 4 x 6
= 6 + 4 x (4 + 2)


So term n would be

= 6 + n x (n + 2)
= 6 + n² + 2n
= n² + 2n + 6


When the nth term = 446,

n² + 2n + 6 = 446

n² + 2n - 440 = 0

(n - 20)(n + 22) = 0

n = 20
or
n = -22(rejected as n cannot be negative)


So n = 20

Hard to see the pattern at times though.

What I would do is to recognise that the difference of the difference of the terms are fixed, so the pattern must be of the form An2 + Bn + C = number.

Then sub in the first three numbers of the pattern and proceed to solve for A, B and C using the three equations formed.

Eric nicholas Sir , can you please help me with the recent question I posted .thank you a lot ;)

Another way :

Term 1 = 9
= 4 + 5
= 4 + 5 x 1 + 0
= 4 + 5 x 1 + (1 - 1)(1 - 2)

Term 2 = 14
= 9 + 5
= 4 + 5 x 1 + 5
= 4 + 5 x 2
= 4 + 5 x 2 + 0
= 4 + 5 x 2 + (2 - 1)(2 - 2)


Term 3 = 21
= 14 + 7
= 4 + 5 x 2 + 5 + 2
= 4 + 5 x 3 + 2 x 1
= 4 + 5 x 3 + (3 - 1)(3 - 2)

Term 4 = 30
= 21 + 9
= 4 + 5 x 3 + 2 x 1 + 5 + 2 x 2
= 4 + 5 x 4 + 3 x 2
= 4 + 5 x 4 + (4 - 1)(4 - 2)


Term 5
= 30 + 11
= 4 + 5 x 4 + 3 x 2 + 5 + 2 x 3
= 4 + 5 x 5 + 4 x 3
= 4 + 5 x 5 + (5 - 1)(5 - 2)

Term 6
= 30 + 11 + 13
= 4 + 5 x 5 + 4 x 3 + 5 + 2 x 4
= 4 + 5 x 5 + 5 x 4
= 4 + 5 x 5 + (6 - 1)(6 - 2)


So ,

Term n

= 4 + 5 x n + (n - 1)(n - 2)
= 4 + 5n + n² - 3n + 2
= n² + 2n + 6


The rest is the same as above

At sec 2 such questions are given to test their ability to see such patterns/hone the skill.

Your method is more suitable for H2/A levels (common difference / common difference of differences). At sec 2, it's even harder to fathom how one gets to the formula an² + bn + c

There is a guideline for this though. I was taught this approach in Sec 3 if I recall.

I have included the guideline in my posted workings.

There is yet a different approach purely using basic observation of a common, familiar pattern.

Strangely enough, never had this guideline appeared during my time.

But somehow it works for powers one and two. I believe it should work for higher powers as well though, but have not actually experimented on these in practice.

Mr eric , I posted one last question that I have difficulties with . If you dont mind , please help to answer the question. Thanks once again

Mr J , if you are free , please also try to solve my latest question , having difficulties . Thanks a lot

It should work for all positive powers since it's based on derivatives.


For power 2 (2 layers), differentiating an² + bn + c twice gives the constant 2a

(which is the constant difference of differences you mentioned)

The working is similar for other layers.

I see. thank you teachers for explaining. Please try to help me with my latest question posted

Okay