proof by induction markscheme

[thin ivy]

markscheme doesnt explain this step?

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[thin ivy]

for the series

its meant to be (k+1) btw the print just didnt work idk why

[wind mauve]

Ok so notice that the 3^0 when added just extends the series such that instead of being
3^1 + 3^2 + 3^3... + 3^k
it becomes
3^0 + 3^1 + 3^2.. + 3^k
so we started at r = 0 instead of r = 1
So (the sum from r = 1 to k of 3^r) + 3^0 can be simplified to
the sum from r = 0 to k of 3^r
now what they do is just a substitution by adding 1 to the lower and upper bound (replacing r = 0 with r = 1, and k with k+1)
this turns the series from
3^0 + 3^1 + ... + 3^k
to
3^1 + 3^2 + ... + 3^(k+1)
to accoount for this they turn 3^r into 3^(r-1) so one is subtracted from every term so it becomes
3^(1-1) + 3^(2-1)... + 3^(k+1-1)
which just simplifies to
3^0 + 3^1 + ... + 3^k our original series

[thin ivy]

i also understand up to “the sum from r=0 to k of 3^r”

i dont understand from there after

what do u mean substitution? sry im confused

[wind mauve]

if we originally had the sum from r = 0 to k of 3^r it would be
3^0 + 3^1 + 3^2... + 3^k
if we changed the bounds by adding 1 to each bound, it would become
the sum from r = 1 to k+1 of 3^r which is
3^1 + 3^2 + ... + 3^(k+1)
clearly this is different from the previous one
so we need to account for the change we've made, notice that if we decrease the power of 3 in all of them by 1 it returns us back to the original series since
3^(1-1) + 3^(2-1) + 3^(3-1)... + 3^(k+1-1) = 3^0 + 3^1 + 3^2... + 3^k
so we just replace 3^r with 3^(r-1) and that makes our new bounds work,
so the sum from r = 0 to k of 3^r = the sum from r = 1 to k+1 of 3^(r-1)
generally speaking,
the sum from r=x to r=y of f(r) = the sum from r = x+n to r = y+n of f(r-n)
because when we add the constant n to both bounds we just add that constant to the input of each term in the series, so when we immediately subtract it (by doing f(r-n) instead of f(r)) we end up with the same series as the original
Sorry my explanation is kind of all over the place

[clear jasper]

Okay, this is just really weird indexing. I would personally leave it like $\sum_{r=0}^k 3^r$, but they did a “change of index” and “shifted” everything to the right by 1 for some reason; i.e., every letter involved in the indexes is increased by 1, and the letter in the summand decreases by 1.

[manic sparrow]

Kocher

[clear jasper]

Think about this; you still start with 3^0 (3^(1-1)=3^0), and end with 3^k (3^((k+1)-1)=3^k).

In general, $\sum_{r=0}^k a_r=\sum_{r=i}^{k+i}a_{r-i}$.

[manic sparrow]

Kocher

[clear jasper]

In this case, i=1.

[thin ivy]

thanks guys reading over it now

[thin ivy]

i see,so its kind of trial and error

in a way

[thin ivy]

im gonna remember this also, tysm

thank you guys

this was really helpful

+close

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