Math question help

Yes you are right

Thanks very much!
Also, I get how to do this question as well.

Attachment not found

But was just wondering.. Isn’t a_k=2^k the kth term of a geometric sequence because we’re multiplying by 2 each time?

For a_k=2^k to be the kth term of a geometric sequence also doesn’t make sense though, and I have explained this below:

a₀=1a₁ =2a₂ =4
a₃ =8
a₄ =16

We are multiplying by 2 each time. The first term (a₁ =2 not a₀ =1) is 2.

Using the formula a_k =ar^k-1, the kth term of the geometric sequence is a_k =2 x 2^k-1. This doesn’t make sense as the kth term mentioned in the question for this sequence is a_k=2^k..
Could you please help me understand this?

Look at the number below the sigma notation, the first one start with 1, while the second one starts at 1 (1 to 4)

Don't understand your question TBH.

Of a geo sequence, $ar^{k}$ does not specify the kth term, it specifies the (k+1)th term. But $ar^{k-1}$ does specify the kth term, and note that $2 \cdot 2^{k-1} = 2^k$

Ahh this makes sense cause 2 x2^(k-1) is the same thing as 2^k!

Just to double check:
2iii) is a geometric series, right?
And the first term of this geometric series is a1=2 and not a0=1 right?

1(iv) is a geometric series as well?
And the first term of this geometric series is a1=1?

So a1 is pretty much the first term for any geometric/arithmetic series/sequence?

It's a geometric series of the sequence $a_k = 2^{k-1}$ with $a_1=1$.



It's a geometric series of a sequence $a_k = (-2)^{k-1}$ with $a_1=1$



Yes, you want $a_1$ to correspond to the 1st term of your sequence. Wouldn't make much sense to refer to the '0th' term

Thanks, but isn't 2iii) ak = 2^k or ak=2 x 2^(k-1)? Also, shouldn't a1 = 2?
I have just attached it here again:

If it starts at 2, then you're completely ignoring the actual first term of 1.

Note that $\displaystyle \sum_{k=0}^{k=4} 2^k = \sum_{k+1=1}^{k+1=5} 2^{(k+1)-1}$ then map $(k+1) \mapsto k$ and we have $\displaystyle \sum_{k=1}^5 2^{k-1}$ which is the sum of the sequence $\{ 2^{k-1} \}$ from the first term up to the 5th term.

...

In that case the formula is just $a_k=2^k$ if you start from $a_0$ as the initial term.

But the thing is if you start from a0 as the initial term, the kth term of the sequence is ak=2^(k-1).

This is shown here:
r=2 and a0=1
Sub them both into the formula ak=ar^(k-1) and you get ak=2^(k-1) and not ak=2^k.

In the picture I have attached here, ak=2^k where a0=1,a1=2,a2=4,a3=8 and a4=16
so why is it only when we sub a1=2 into ak=ar^(k-1), that we get ak=2^k?
As you said, "the formula is just if you start from as the initial term" but this is not the case here.
Would really appreciate if you could please let me know what you think. Many thanks.

If you decide to allow the 0th term as a thing, then the sequence is $\{ 2^k \}$ for $k=0,1,2,...$ so that $a_0=1, a_1=2, ...$ and we can say that the kth term here is $a_k=2^k$ but we have to start with 0th term. So when you say 2nd term, you're really referring to the 3rd one down the line.