In this video we're talking about how to take a sum and rewrite it in summation
notation. And in this particular problem we've been given this sum here, so one
half plus one fourth plus one 8th one 16th one 32nd and one over 64 164th. What we need to do
is you summation notation to simplify this sum. So when we talk about
summation notation we mean Sigma notation. So we have this Sigma value
here with n equal to something. We're going to have a number here on top of the
Sigma notation and we're going to need to find a function that represents this
sum. So this is a tricky thing to learn how to do but basically what it comes
down to is you're just looking for patterns and sometimes they're going to be
more complicated than others. In this case we're going to do a really simple one
so that we can get the feel for it. But all we're doing is looking for patterns.
So if we look at this sum, the first thing we notice is every one of these
terms is a fraction and the numerator of each of the fractions is one so it's a
pretty good bet that the function that represents this sum is going to be a
fraction where the numerator is one. So let's just go ahead and write that. The
denominator is a little bit trickier because the number changes each time. But
what we see is that they're all multiples of 2 in fact they're all powers
of two. So we could actually rewrite these denominators. We could call this
one two the first power. We could call this one two scared because two
squared is four. We know that to the third is eight we could say this is to the
third. We know that to the fourth power is 16 so this is two the fourth. 32 is two
the fifth and 64 is two to the sixth power. So since that's the case the denominator of our
function is probably going to have base two. The only issue is that the exponent here
always changes. Notice it's counting up 1, 2, 3, 4, 5, 6. So how do we deal with that if we
want to write a function that represents this sum? Well what you always need to
remember and you should go ahead and write this below each term in your
expanded sum, is that this first term is usually going to correspond to either n equals zero
or n equals one. In this case it's going to be n equals one and I'll show you why
in a second. So if we say this is going to be n equals one, this is going to be n
equals two, n equals three, n equals four, n equals five and n equals six. You always
just count up from either one or zero it's just really matter of whether or
not you're starting with n equals one or n equals zero. But wherever you start, you
count up from that number one integer at a time. But now if we say that, for this
term n is equal to one,
well, this is one here, so that could be n. Here if n is equal to two, this is
two, so that could be n. Here n is equal to three and this exponent is three so
that could be n. So what we see is that if we assign values n equals 1, 2, 3, 4, 5, 6, that correspond
with each of these terms, then what this really becomes is one over two to the n,
and one over two the n
can represent each of these terms as long as n starts at one. If we had said
n starts at zero. If we've started it at n equals zero and then done n equals 1, 2, 3, 4
and 5, then this first exponent right here would have been n plus one because
this would have been the n equals 0 term and to get to one here we would have had to add
one so it would have been the same as n plus one. And we could have done that we could
have said n equals zero and then the function would have been one over two to the
n plus one, but it's cleaner just to say n equals one, so that we can start
this at one over two the n. And then what we can say is, since this series ends at n equals six, we'll say
the index is n equals one, up to six, and now this is summation notation that
represents this series and if we want to double check ourselves, all we have to do
is plug this back in. So here we start at n equals one and we know we're going to
go up to six. But if we plug in n equals one first, what we're going to get is one over two
the first power or one over two. If we then start counting up and we say n equals
two, we plug in n equlas two, we're going to get one over two squared or
one-fourth. Plugging in n equals three we're going to get two cubed or eight so we
get one 8th. And we could keep going up to n equals six. If we plug in n equals
six we're going to get one over to the sixth power or one over 64. And what you
can see is that this expanded sum is the same as the expanded sum we were
given in our original problem. We start at one half, we go up six terms and we end at
one over 64 and all the terms in the middle are going to match these original
terms. What that tells us is that we did this correctly and that one over two to the
n does represent this expanded sum. And so we can say then that this is our
final answer here and this is the summation notation of the original sum.
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