Summation Formulas and Rules
n
X
“Adding c to itself n times
same as multiplying n and c.”
c = n·c
k=1
n
X
c · ak = c ·
n
X
k=1
n
X
Generalization of
c · a + c · b = c · (a + b)
ak
k=1
(ak ± bk ) =
k=1
n
X
ak ±
k=1
m
X
n
X
bk
Generalization of
(a±b)+(c±d) = (a+c)±(b+d)
k=1
ak +
k=1
n
X
ak
=
k=m+1
n
X
ak
k=1
Sums of Powers
n
X
k=1
k
=
n(n + 1)
2
=
1 2
1
·n +
·n
2
2
Sum of first n integers is ≈ half the square of the nth integer.
n
X
k=1
k2
=
n(n + 1)(2n + 1)
6
=
1 3
1 2
1
·n +
·n +
·n
3
2
6
Sum of first n squares is ≈ one third of the cube of the nth
integer.
Sums of Powers (continued)
n
X
k
3
n2 (n + 1)2
4
=
k=1
=
1 4
n + Lower order terms in n
4
Sum of first n third powers is ≈ one fourth of fourth power of n.
n
X
k
4
=
k=1
n(n + 1)(2n + 1)(3n2 + 3n − 1)
30
1 5
n +
5
=
Lower order terms in n
Sum of first n fourth powers is ≈ one fifth of fifth power of n.
Example 1
Evaluate the sum
9
X
(5k + 8)
k=1
9
X
(5k + 8)
=
5·
k=1
9
X
k
+
k=1
=
5·
9 · 10
2
=
+
297
9
X
k=1
9·8
8
Example 2
Evaluate the sum
8
X
(5k 2 + 8k + 1)
k=1
=
5·
8
X
k
2
8·
+
8
X
k=1
=
5·
k
+
k=1
8 · (8 + 1) · (2 · 8 + 1)
6
+
=
8
X
1
k=1
8·
8 · (8 + 1)
2
8·1
+
1316
Example 3:
Evaluate the sum
12
X
(k + 1)
k=7
12
X
(k + 1)
=
12
X
k=7
k=1
=
−
k=1
12
X
=
(k + 1)
k +
12
X
−
1
k=1
6
X
k=1
=
(k + 1)
k=1
!
12 · (12 + 1)
+ 12
2
6
X
−
63
k +
6
X
!
1
k=1
6 · (6 + 1)
+ 6
2
Example 3 (Alternate solution)
Notice
12
X
(k + 1)
13
X
=
k=7
=
k=8
13
X
−
k
k=1
=
k
7
X
k
k=1
13 · (13 + 1)
7 · (7 + 1)
−
2
2
=
63
Example 4:
Evaluate the sum
100
X
(2 + 5k)
k=3
100
X
(2 + 5k)
=
k=3
But
100
X
(2 + 5k)
−
k=1
100
X
(2 + 5k)
=
2 · 100 + 5 ·
100
X
k=1
200
+
=
(2 + 5k)
k=1
k=1
=
2
X
5·
100 · (100 + 1)
2
25250
k
Example 4 (continued)
Also
2
X
(2 + 5k)
(2 + 5 · 1) + (2 + 5 · 2)
=
=
k=1
So
100
X
(2 + 5k)
25250 − 19
=
=
25431
k=3
Example 4 (alternate solution)
Notice
100
X
(2 + 5k)
98
X
and
k=3
(12 + 5k)
k=1
both represent the sum
17 + 22 + 27 + · · · + 502
But
98
X
(12 + 5k)
12 · 98
=
k=1
+
5·
98
X
k=1
=
12 · 98
=
+
5·
98 · (98 + 1)
2
25431
k
19
Example 5:
Evaluate the sum
1 + 5 + 10 + 15 + 20 + · · · + 245
What is the pattern? The +1 seems to get in the way.
Group terms in the following way
1 + (5 + 10 + 15 + 20 + · · · + 245)
=
1 + 5 · (1 + 2 + 3 + 4 + · · · + 49)
=
1 + 5·
49 · (49 + 1)
2
=
6126
Example 6
Evaluate the sum
24 + 27 + 30 + 33 + 36 + · · · + 90
=
=
3 · (8 + 9 + 10 + · · · + 30)
3 · (1 + 2 + 3 + · · · + 30) − 3 · (1 + 2 + 3 + · · · + 7)
=
3·
30 · (30 + 1)
7 · (7 + 1)
− 3·
2
2
=
1311
Example 7
Evaluate the sum
−5 − 4 − 3 − 2 − 1 + 0 + 1 + 2 + 3 + · · · + 30
=
−(5 + 4 + 3 + 2 + 1) + 0 + (1 + 2 + 3 + · · · + 30)
−
=
5 · (5 + 1)
30 · (30 + 1)
+ 0 +
2
2
=
450
Example 8
If we write
n
X
k
4
4n(n + 1)(2n + 1)(3n2 + 3n − 1)
,
A
=
k=1
what is the value of A?
Let n = 1.
Left hand side
Right hand side
=
=
14
=
1
4 · 1(1 + 1)(2 · 1 + 1)(3 · 12 + 3 · 1 − 1)
A
=
120
A
Example 8 (continued)
Equating left and right hand sides, get
1
=
120
A
A
=
120
so
Example 9
If we write
n
X
(k 2 − k) =
k=1
2n(n + 1)(n − 1)
,
A
what is the value of A?
Plugging in n = 1 makes both sides zero, which gives no
information.
Try n = 2:
Left hand side = (12 − 1) + (22 − 2) = 2
Right hand side =
2 · 2(2 + 1)(2 − 1)
12
=
A
A
Equating left and right hand sides:
2=
12
A
=⇒
A=6
Key steps in Computing a Definite Integral
We want to compute definite integrals,
Z b
f (x) dx
a
which were defined as limits of sums of areas of rectangles:
lim
n
X
n→∞
f (pk ) ∆xk
k=1
First: find the various f (pk )’s. (See Chapter 8)
Next: use formulas from this chapter to evalute the sum
n
X
f (pk )∆k
k=1
Finally: use techniques from Chapter 3 to compute the limit.
Next few examples review certain limits.
Example 10:
Find
8n2 + 7n + 9
lim
n→∞ 4n2 + 2n + 1
For large n:
Numerator
Denominator
≈ 8n2
≈ 4n2
So for large n,
8n2 + 7n + 9
8n2
≈
4n2 + 2n + 1
4n2
=
2
In the limit, the approximation becomes exact:
8n2 + 7n + 9
lim
n→∞ 4n2 + 2n + 1
=
8n2
lim
n→∞ 4n2
=
2
Example 11 :
Find
(2n + 1)2
lim
n→∞ 5n2 + 2n + 1
For large n,
2n + 1 ≈ 2n
so
(2n + 1)2 ≈ (2n)2 = 4n2
Also,
5n2 + 2n + 1 ≈ 5n2
So
(2n + 1)2
lim
=
n→∞ 5n2 + 2n + 1
4n2
4
lim
=
n→∞ 5n2
5
Example 12:
Find
n4 + n2 + 13
lim
n→∞ n3 + 8n + 9
For large n,
Numerator
Denominator
so
≈ n4
≈ n3
n4 + n2 + 13
n4
≈ 3 = n
n3 + 8n + 9
n
So
n4 + n2 + 13
lim
n→∞ n3 + 8n + 9
=
lim n
n→∞
=
DNE
Signed Areas
Riemann Sums and Definite Integrals were introduced to
compute area under a curve.
But something funny happens if y = f (x) is negative:
Suppose f (x) = −2.
Right hand sum on [0, 2] with two subdivisions uses
∆x = 1
x1 = 1,
x2 = 2
f (x1 ) = −2,
f (x2 ) = 2
So sum is
2
X
f (xk )∆x
(−2) · 1 + (−2) · 1 = −4
=
k=1
How can this sum represent an area? Aren’t areas supposed to
be postive?
Signed Areas
It doesn’t matter how many subdivisions we use, or whether we
use left or right sums, any approximating sum for the integral
of f (x) = −2 will always be negative.
Riemann Sums and Definite Integrals compute net area, or
signed area. By this, we mean that area below the x-axis is
counted with a negative sign.
In general,
Z
b
f (x) dx
a
= (area or regions above x-axis )−( area or regions below x-axis)
Velocity, Distance, Displacement
Likewise, the integral of a velocity is not quite the distance
travelled. This is correct only if the object is always travelling
in the same direction.
Throw a ball straight up in the air with initial velocity 64 feet
per second.
The velocity at time t is then
v(t) = 64 − 32t
Then
Z
4
v(t) dt = 0
0
!? The distance travelled is 0 ?!
The ball did travel. It travelled up, then it travelled down; it
ended up exactly where it started.
Velocity, Distance, Displacement
The integral of the velocity of an object is the displacement of
the object.
The displacement is the change in position:
displacement
=
final position − initial position
If the velocity is always positive (object always moves in same
direction) the displacement coincides with distance travelled.
If the velocity changes signs (object moves in one direction,
then comes back) the displacement will be less that the distance
travelled.
Displacement may be positive or negative, where as distance is
always positive.
Example 13
Evaluate the limit as n tends to infinity:
n
1 Xk+9
lim
n→∞ n
n
k=1
First, simplify the sum:
n
X
k+9
k=1
n
n
1 X
1
=
(k + 9) =
n
n
k=1
=
1
n
n
X
k +
k=1
n · (n + 1)
+ 9·n
2
n
X
k=1
n+1
+ 9
2
=
n + 19
2
=
Example 13 (continued)
Thus,
n
X
k+9
n
k=1
and so
=
n
1 Xk+9
=
lim
n→∞ n
n
k=1
=
n + 19
2
lim
n→∞
lim
n + 19
2n
=
1
2
n→∞
1 n + 19
n
2
!
9
Example 14
Evaluate the limit as n tends to infinity.
n 1 X
7 2
lim
k·
n→∞ n
n
k=1
First, simplify the sum
n 1 X
7 2
k·
n
n
=
k=1
=
n
1 X 2 72
k · 2
n
n
k=1
n
49 X 2
k
n3
k=1
49 n(n + 1)(2n + 1)
n3
6
=
=
49 (n + 1)(2n + 1)
6n2
For large n,
49 (n + 1)(2n + 1)
49 (n)(2n)
≈
6n2
6n2
=
So
49
3
n 7 2
1 X
lim
k·
n→∞ n
n
k=1
=
49
3
Example 15:
The integral
R5
0
x2 dx is computed as the limit of the sum
n
X
A 2
A
k
n
n
k=1
What value of A must appear in the sum?
a = 0, b = 5, so
∆x =
b−a
5
=
n
n
The partition points are of the form
xk = a + k · ∆x = k ·
5
n
Example 15 (continued)
The right hand sum is then
n
X
∆x (xk )2
k=1
=
n
X
5 2
5
k
n
n
k=1
Thus,
A
n
and so A = 5.
A 2
5
5 2
=
k
k
n
n
n
Example 16 :
The integral
R 10
7
x2 dx is computed as the limit of the sum
n
X
3
3 2
A+k·
n
n
k=1
What value should be used as A?
a = 7 and b = 10 so
∆x =
b−a
3
=
n
n
The right endpoints are given by
xk = a + k · ∆x = 7 + k ·
The right sum is then
n
X
3 2
3
7 + k·
n
n
k=1
Comparing the right hand sum to
n
X
3
3 2
A + k·
n
n
k=1
we see A = 7.
3
n
Example 17
The limit
n
X
3 n+k 2
lim
n→∞
n
n
k=1
is obtained by applying the definition of the integral to
Z 2
f (x) dx
1
What is the function f (x)?
a = 1, b = 2 so
b−a
1
=
n
n
∆x =
The right endpoints are of the form
xk = a + k · ∆x = 1 + k ·
1
n+k
=
n
n
Example 17
The right hand sum is then
n
X
∆x f (xk )
n
X
1
n+k
f
n
n
=
k=1
k=1
Equating the summands, find
1
f
n
n+k
n
=
3
n
n+k
n
so
f
Thus, f (x) = 3x2 .
n+k
n
=
3
n+k
n
2
2
Example 18:
Evaluate the limit
2
lim
n→∞ n
n
X
k=1
s
2
4− k·
n
2
Hint: Think
about how this limit relates to the area under
√
f (x) = 4 − x2 from x = 0 to x = 2.
a = 0, b = 2, so
∆x =
2
b−a
=
n
n
The right endpoints are
2
n
xk = a + k · ∆ = k ·
The right hand sum is then
n
X
q
∆x 4 − x2k
k=1
=
n
X
2
n
k=1
Taking the limit, we see
s
n
2 X
2 2
lim
4− k·
n→∞ n
n
k=1
s
2
4− k·
n
Z
=
2
2
p
4 − x2 dx
0
√
R2 √
But 0 4 − x2 dx is the area under the curve y = 4 − x2
from x = 0 to x = 2, and
p
y = 4 − x2
=⇒
x2 + y 2 = 22 ,
Area we are interested in is the area of a quarter circle of radius
2.
Area of full circle is π · 22 = 4π.
The area of the quarter circle of radius 2 is then
1
· 4π
4
=
π
Thus,
n
2 X
lim
n→∞ n
k=1
s
2 2
4− k·
n
=
π
Example 19:
Given the area of the ellipse 49x2 + y 2 = 49 is 7π, evaluate the
integral
Z 1p
49 − 49x2 dx
0
Hint: Think of integral as an area.
√
Let y = 49 − 49x2 . This rearanges to 49x2 + y 2 = 49. Now,
p
y =
49 − 49x2 ≥ 0
so integral is only concerned with the area above the x-axis.
Also, limits of integration imply integral is only concerned with
area to the right of y-axis.
Z 1p
49 − 49x2 dx = Area of part of ellipse in first quadrant
0
=
1
· Area of Ellipse
4
=
7π
4
Example 20
A car is travelling due east. Its velocity (in miles per hour) at
time t is given by
v(t) = −2.5t2 + 10t + 50
How far did car travel during the first five hours of the trip?
If velocity was constant, then could use
Distance Travelled = Velocity · Time Elapsed
But velocity is changing, so need to think of distance as area
under velocity curve.
Z 5
Distance Travelled =
(−2.5 t2 + 10t + 50) dt
0
Example 20
We could compute this distance by simplifying a Riemann Sum,
then taking a limit.
However, the calculation would be very long and tedious.
We discuss a more clever method.
Recall
Distance 0 = Velocity
This was used when distance was given and we wanted velocity.
Lets try using it in reverse! Can we find a distance function,
D(t) whose derivative is the given velocity
D0 (t) = −2.5 t2 + 10t + 50?
Derivatives reduce powers by one, so D(t) should involve t3 , t2 ,
and t. Correcting for constants, we see
D(t) = −
2.5 3
t + 5t2 + 50t
3
satisfies
D0 (t) = −2.5 t2 + 10t + 50.
How did we come up with D(t)? More on that in Chapter 10.
Lets just see how this helps solve the current problem.
3
2
Now, − 2.5
3 t + 5t + 50t is not the only function whose
derivative is equal to v(t). For instance,
0
2.5 3
2
−
t + 5t + 50t + 183692 = −2.5 t2 + 10t + 50
3
But a consequence of the Mean Value Theorem (Chapter 6)
assures us that any such D(t) will be
D(t) = −
for some constant C.
2.5 3
t + 5t2 + 50t + C
3
How to find the correct value of C?
How far did the car go in 0 hours? It went nowhere, so
D(0) = 0. Plug in to find C:
0 = D(0) = −
2.5 3
· 0 + 5 · 02 + 50 · 0 + C = 0 + C
3
Thus, C = 0 and the distance travelled after t hours is
D(t) = −
2.5 3
t + 5t2 + 50t
3
and so the distance travelled after 5 hours is
D(5) = −
2.5 3
5 + 5 · 52 + 50 · 5 = 270.83
3
If you don’t like this argument, you can try computing the
integral with Riemann Sums:
Example 20 (with Reimann Sums)
Have a = 0, b = 5 and so
b−a
5
=
n
n
Right Endpoints are of the form
∆x =
tk = a + k · ∆t = k ·
5
5k
=
n
n
Right hand sum is then
n
X
k=1
n
X
5
∆t v(tk ) =
(−2.5 t2k + 10tk + 50)
n
k=1
2
n
X
5
5k
5k
=
(−2.5
+ 10
+ 50)
n
n
n
k=1
!
n
2
X
5
62.5k
50k
=
(−
+
+ 50)
n
n2
n
k=1
!
n
n
62.5k 2 X 50k X
−
+
50
+
n2
n
k=1
k=1
k=1
!
n
n
X
X
5
62.5
50
=
− 2
k + 50n
k2 +
n
n
n
5
=
n
n
X
k=1
k=1
5
62.5 n(n + 1)(2n + 1) 50 n(n + 1)
=
− 2 ·
+
+ 50n
n
n
6
n
2
5
62.5 (n + 1)(2n + 1)
=
·
+ 25(n + 1) + 50n
−
n
n
6
5
62.5(n + 1)(2n + 1)
=
−
+ 25(n + 1) + 50n
n
6n
=−
312.5(n + 1)(2n + 1) 375n + 125
+
6n2
n
For large n, this is
≈−
312.5(n)(2n) 375n
312.5
+
=
−
+ 375 = 270.83.
6n2
n
3
Car traveled 270.83 miles in five hours.
Does this answer seem reasonable? First, find max and min
velocity over the 5 hour period. Endpoints: t = 0 and t = 5.
Critical points: v 0 (t) = −5t + 10 = 0 so t = 2. Plug in:
v(0) = 50
v(2) = 60
v(5) = 37.5
Minimum velocity is 37.5. Thus, car travels at least
5 · 37.5 = 187.5 miles. Maximum velocity is 60. Thus, car travels
no more that 5 · 60 = 300 miles. Thus, rough approximation is
that car travels somewhere between 187.5 and 300 miles. This
is consistent with the answer we got from the long calculation.