MATH 101B ASSIGNMENT 1
1. (a) Using the picture below, calculate the sum
5
X
i.
i=1
(There are many ways of calculating this sum, including directly. You must use the picture and
explain in a few sentences how you have done so.)
In this question, we are asked to calculate the sum of
5
X
i. Considering the picture, we could
i=1
interpret the given white block as being half of the entire rectangle. Therefore
(6blocks ∗ 5blocks)/2 = 15whiteblocks
or the given white block is half of the total rectangle.
(b) Use the idea from part (a) to come up with a formula for
n
X
i, where n is any positive integer.
i=1
Considering the rationale behind the previous question, we can generalize an equation for the sumn
X
mation of
i for any particular value of n. The last questions answer shows us that
i=1
5
X
i = (6blocks ∗ 5blocks)/2 = 15whiteblocks
i=1
or
(5 + 1)(5)/2 = 15.
We can then surmise that ((n+1)(n))/2=
n
X
i, which is our final answer.
i=1
2. (a) Since we now know how to calculate
n
X
i, we can calculate
i=1
n
X
i2 . For this, first consider the
i=1
difference
i3 − (i − 1)3 = 3i2 − 3i + 1.
Then, come up with a formula for
n
X
i=1
i2 by summing both sides of (1) from 1 to n.
(1)
To find a formula for
n
X
i2 we begin by summing both sides of the given difference i3 − (i − 1)3 =
i=1
n
X
2
3i − 3i + 1 and then isolating for
i2 . The initial summation is the formula is
i=1
n
X
n
X
(3i2 − 3i + 1).
i=1
i=1
(i3 − (i − 1)3 ) =
The left-hand side of the equation can be understood as an integer subtracted from the previous
integer, therefore, due to the nature of this summation, a sum to the n’th value would be n3 as all
previous integers would be canceled out by a prior term. Understanding this, the equation
n3 = 3 ∗
n
X
i2 − 3 ∗
i=1
n
X
i+
i=1
n
X
1
i=1
follows. Then we can plug the known formula from part 1, as well as sum the constant terms with
n
n
X
X
the rule of: n ∗ constant =
C, and isolate for
i2 . These operations result in the equation
i=1
i=1
3∗
n
X
i2 = 3 ∗ (((n + 1)(n))/2) − n + n3 .
i=1
The equation can be simplified to
(n(2n2 + 3n + 1))/6 =
n
X
i2
i=1
which is our final answer.
(b) Use a similar argument1 as in 2(a) to find a formula for
n
X
i3 .
i=1
We can utilize an augmented version of the argument from the previous part to solve for
n
X
i3 .
i=1
Following this, the formula
i4 − (i − 1)4 = 4i3 − 6i2 + 4i − 1
is found. Summing this equation yields a similar result to the previous part, where every term before
the n’th term will cancel out due to a prior term, showing
n
X
i4 − (i − 1)4 = n4 .
i=1
We then set the sum and split the sum of the left-hand side equal to the sum of the right-hand side
n
X
and isolate for
i3 . This equation will look like this:
i=1
4∗
n
X
i3 = 6 ∗
i=1
1 In
n
X
i=1
i2 − 4 ∗
n
X
i=1
i+
n
X
1 + n4 .
i=1
fact, this method is generalizable. One can also show that if we know the formulas for
n
X
ik for 1 ≤ k ≤ m (k,m are
i=1
positive integers), we can calculate a formula for
n
X
i=1
do here).
i
m+1
. This involves the more general binomial expansion (which we won’t
Plugging in our found equations for
n
X
2
i and
n
X
i=1
i and utilizing the constant rule for summation
i=1
defined in the previous question gives
4∗
n
X
i3 = ((6n)(2n + 1)(n + 1))/6 − ((4n)(n + 1))/2) + n + n4 .
i=1
Simplifying and dividing by four gives us our final answer
n
X
i3 = (n2 (n + 1))/4.
i=1
n
X
2
3. (a) Use one or more of the formulas derived above to calculate lim
n→∞
n
i=1
!
2
i
+1 .
n
We start by distributing (2/n) into the brackets
n 2 X
2
2i
+
lim
.
n→∞
n3
n
i=1
Then split the summation into parts and pull out
2
n3
and
2
n
n
n
2 X 2
2X
i
+
lim
(1) .
n→∞ n3
n→∞ n
i=1
i=1
lim
We plug in the found formula for
n
X
i2 and use the summation constant rule defined above yielding
i=1
lim
n→∞
2 n(n + 1)(2n + 1)
2n
+ lim
.
∗
n→∞ n
n3
6
Simplifying we find:
2n3 + 3n2 + n
+ 2 lim 1.
n→∞
n→∞
3n3
Evaluating the limit where only the highest n powers will contribute yields
lim
8
2
+2=
3
3
which is our final answer.
Z
1
(b) The limit in part (a) represents an integral
f (x) dx. What is f (x)? Remember to justify your
0
answer carefully.
Use the formula of the Riemann sum for this problem. That equation is:
X
lim
f (xi) ∗ ∆x.
n→∞
i=1
∆x = b−a
n where b is the upper bound and a the lower bound of the given integral. Now set the
equation in part a to the sum we created with the equation as follows:
!
2
n
n
X
X
2
i
+ 1 = lim
f (a + i∆x) ∗ ∆x.
lim
n→∞
n→∞
n
n
i=1
i=1
Now we substitute the
1
n
for the ∆x and the a value of 0 in:
!
2
n
n
X
X
2
i
1
1
lim
+ 1 = lim
f (0 + i ) ∗ .
n→∞
n→∞
n
n
n
n
i=1
i=1
Evaluating and simplifying this equation using the same we get:
i
i
2 ∗ (( )2 + 1) = f ( ).
n
n
Finally we substitute
i
n
= x and find the equation
2x2 + 1 = f (x).