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ADDITIONAL EXERCISES
SOLUTIONS
Let n be any natural number. Then, the value of 1 + 2 +. . . +(n -1) + n equals n(n+1)/2.
A simple proof of this fact follows: Call s = 1 + 2 +...+ (n -1) + n ; thus, we need to
compute the value of s.
Observe that (obviously) s = n+(n-1) +...+ 3+2+1, because we rewrote the sum, in
reversed order. Therefore, 2s= s+s={1+ 2 +...+(n -1) + n} + {n + (n-1) +...+ 3 + 2 +1}. To
add, "match" each number in the first bracket with the corresponding one in the second
bracket. So, 1 with n, 2 with (n-1), 3 with (n-2), etc... Each sum equals n+1 and there are
n of those sums. Therefore, 2s = (n) (n+1) and finally s = n (n+1)/2, as claimed.
Note: the result can be proved in many other different ways.
The following problems are based on the formula just obtained:
1 + 2 + . . . + (n –1) + n =
(n)(n + 1)
2
€
1) Write down the first n odd positive integers 2k+1. Use summation sign to express
their sum. Compute the sum of the first n odd positive integers.
SOLUTION: notice first that k must range from 0 to n-1, producing therefore the
sequence 1, 3, . . . , 2n-1. (Make sure to understand this!) Their sum can be written as
n−1
n−1
n−1
∑ (2k + 1) = 2
∑ k +
n
=
2
∑ k +
n
=
k= 0
€
k= 0
k=1
2(n −1)(n)
+
n
=
(n‐1)
n
+
n
=
n 2
2
2) Write down as a summation sign and compute the €sum of the first n even positive
integers,
that is,€the sum of€ 2k for k=1, 2, ..., n. (Hint: can you factor something out?)
€
n
n
SOLUTION:
∑ 2k =
2 ∑ k =
k=1
€
€
k=1
€
2(n)(n + 1)
=
n(n+1)
2
3) Compute the sum of 3k+1, where k ranges from 0 to 100. (Hint: same argument as
before)
100
SOLUTION:
100
100
∑ (3k + 1) =
3 ∑ k +
101
=
3 ∑ k +
101=
k= 0
=101(151)
=
15251
k= 0
k=1
3(100)(101)
+
101
=
101
(150
+
1)
2
Notice:
101
appears
because
we
are
adding
101
times
the
number
1.
€
€
€
€
4) Compute the sum of all consecutive integers from 11 to 2,000. (Hint: we seem to be
lacking some numbers here...)
SOLUTION: call S the sum we must compute. Observe next that (1 + 2 + . . .+10) + S
=1+ 2 + . . .+ 1999 + 2000=
(2000)(2001)
= 2001000. Therefore, solving for S we have
2
S = 2001000 - (1 + 2 +€. . . +10) = 2001000 - 55= 2000945
5) Find the sum of the 100th power of each root of the polynomial x100 - 4x + 5. (Hint:
do not compute the roots. Think what it means for a to be a root)
SOLUTION: if a is a root, then a100 = 4a - 5. Thus we have to compute
100
∑ (4a
j
− 5) =
j=1
100
4 ∑ (a j ) - 500. Since the sum of all the roots equals minus the coefficient of x99 (*),which
j=1
in this case is zero, we have that the answer is - 500
€
€
Note: please accept the claim (*), which can be proved by induction.
After seeing combinatorials solve the following exercises:
6) Write 1 + 2 +...+ (n –1) + n as a single combinatorial.
SOLUTION: 1 + 2 +...+ (n –1) + n =
€
(n)(n + 1)
= nC2
2
7) Recall that the coefficient of qt in (p + q)n is
n
Ct pn-t for all t less than or equal to n.
Compute:
a) the coefficient of x5 in (-3 + x)7
SOLUTION: 7C5 (-3)2 = (9) (21) = 189 since p=-3 and q=x
b) the coefficient of x4 in ( 2 – 3x)16
SOLUTION: 16C4 (2)12 (-3)4 =1820 × 4096 × 81
since p=2 and q=-3x
c) the coefficient of a8 in (2 + 5a2)10
SOLUTION: here p = 2 and q = 5a2 which shows that we must look for the coefficient of
q4. Accordingly, we have 10C4 (2)6 (5)4 = 210 × 64 × 625 = 8,400,000
d) the coefficient of a16 in (2 + 5a2)14
SOLUTION: here p = 2 and q = 5a2 which shows that we must look for the coefficient of
q8. Accordingly, we have 14C8 (2)6 (5)8 = 210 × 64 × 390625 = 5,250,000,000
_______________________________________________________________________
Dr. J. Viola-Prioli