..READ THIS
FIRST..Problems
about Sequences
READ THIS
FIRST
If a problem is solved. It is not 'the' answer.No attempt is made to search for the most elegant answer.
I highly recommend that you at least try to solve the problem before you read the solution.
Problems about Sequences
Level 1 problems
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The angles of a rectangular triangle are the terms of an arithmetic sequence. Calculate these angles.
The angles are (pi/2 - 2h, pi/2 - h, pi/2).
The sum of these angles is 3.(pi/2) - 3.h = pi .
Hence h = pi/6.
The angles are (pi/6, 2pi/6, pi/2) . -
In an arithmetic sequence is t(2) = 3.t(3) .
The sum of n terms, starting from t(1), is 0. Calculate n.The sequence is a, a+v, a+2v, ... a + v = 3.(a + 2v) <=> 2a + 5v = 0 <=> 2a = -5v The n-th term is t(n)=a + (n-1)v The sum of the first n terms is (a + a + (n-1)v).n/2 = 0 <=> (2a + (n-1)v)=0 So, -5v + (n-1)v = 0 and since v is not 0, we have n = 6.
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Calculate 1 + x + x2 + x3 + ... + xn --------------------------------------- 1 + x2 + x4 + x6 + ... + x2nThe numerator is the sum of n terms of a geometric sequence with ratio x. This sum is 1.(xn -1) ----------- (1) (x - 1) The denominator is the sum of n terms of a geometric sequence with ratio x2 . This sum is 1.(x2n - 1) -------------- (2) (x2 - 1) From (1) and (2), we have that the given fraction = (xn -1) (x2n - 1) (xn -1).(x+1) --------- : ------------ = -------------- (x - 1) (x2 - 1) (x2n - 1) -
The sides of a triangle form a geometric sequence. What are the limits of the ratio.
Let a, aq, aqq be the sides of the triangle in ascending order. The condition for the sides isaq2< a + aq <=> q2< 1 + q <=> q2 - q - 1 < 0 <=> q in [1, (1 + sqrt(5))/2 [ -
The points with coordinates (a,b) (a',b') (a",b") are points of a parabola y = 3x2 .
The numbers a, a', a" constitute a arithmetic sequence and b,b',b" form a geometric sequence.
Calculate the ratio of the geometric sequence.Let a = x - h a' = x a" = x + h Then b = 3.(x-h)2 b' = 3.x2 b" = 3.(x+h)2 The condition for geometric sequence gives (3.x2)2 = 9.(x-h)2.(x+h)2 ... h = sqrt(2).x Then the ratio is x2 ------------- = ... = 3 + 2.sqrt(2) (x-h)(x-h) -
Prove that
if a,b,c form an arithmetic sequence, then
b2 + bc + c2, c2 + ca + a2, a2 + ab + b2 form an arithmetic sequence.a,b,c form an arithmetic sequence <=> 2b = a + c Well, b2 + bc + c2, c2 + ca + a2, a2 + ab + b2 form an arithmetic sequence <=> 2(c2 + ca + a2) = b2 + bc + c2 + a2 + ab + b2 <=> c2 + 2ac + a2 = 2b2 + b(a + c) <=> 2c2 + 4ac + 2a2 = 4b2 + 2b(a + c) and since 2b = a + c <=> 2c2 + 4ac + 2a2 = (a + c)2 + (a + c)2 <=> 2c2 + 4ac + 2a2 = 2 (a + c)2 <=> 2c2 + 4ac + 2a2 = 2c2 + 4ac + 2a2
Level 2 problems
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An arithmetic sequence has terms t(1),t(2),t(3),...
The first term t(1) = a and the common difference is v (not 0).
The terms t(5),t(9) and t(16) form a three term geometric sequence with common ratio q.
Calculate q. Calculate t(k) in terms of k and a.t(k) = a + (k-1).v (1) t(5) = a + 4v t(9) = a + 8v t(16) = a + 15v t(5),t(9) and t(16) form a three term geometric sequence a + 4v a + 8v <=> ----------- = ---------- = q a + 8v a + 15v <=> a2 + 19av + 60v2 = a2 + 16av + 64v2 <=> 3av = 4vv Since v is not 0 <=> 3a = 4v <=> v = 3a/4 Combining this with (1) yields t(k) = a + (k-1).3a/4 = (1 + 3k)a/4 Hence t(5) = 4a and t(9) = 7a . So, q = 7/4 -
Prove that for each integer n > 0 1.5 + 2.52+ 3.52+ ... + n.5n= (5 + (4n-1)5n+1)/16We'll prove this by complete induction. a) The property is trivial for n = 1 b) Assume the property is true for n = k. 1.5 + 2.52+ 3.52+ ... + k.5k= (5 + (4k-1)5k+1)/16 Then we have to prove that the property is true for n = k+1. 1.5 + 2.52+ 3.52+ ... + k.5k+ (k+1).5k+1 = (5 + (4k-1)5k+1)/16 + (k+1).5k+1 = (5 + (4k-1)5k+1)/16 + 16.(k+1).5k+1 /16 = (5 + (4k-1)5k+1 + 16.(k+1).5k+1) /16 = (5 + (20k+15)5k+1) /16 = (5 + (4(k+1)-1)5k+2) /16
Level 3 problems
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Calculate the sum of the squares of the first n strictly positive integers. Let S(p) = 1p + 2p + 3p + ... + np . (x + 1)3 = x3 + 3x2+ 3x + 1 Now, make x resp. 0 ; 1 ; 2 ; 3 ; 4 ; ... 13 = + 1 23 = 13 + 3.12+ 3.1 + 1 33 = 23 + 3.22+ 3.2 + 1 43 = 33 + 3.32+ 3.3 + 1 ... (n + 1)3 = n3 + 3n2+ 3n + 1 Adding all these sums together, we have (n + 1)3 = 3.S(2) + 3.S(1) + (n+1) Since S(1) = n.(n+1)/2 (n + 1)3 = 3.S(2) + 3.n.(n+1)/2 + (n+1) 3.S(2) =(n + 1)3 - 3.n.(n+1)/2 - (n+1) 3.S(2) =((n + 1)2 - 3.n/2 - 1 ).(n+1) 3.S(2) =(2.(n + 1)2- 3.n - 2 ).(n+1)/2 3.S(2) =(2n2+ n).(n+1)/2 3.S(2) =n.(n+1)(2n+1)/2 S(2) =n.(n+1)(2n+1)/6
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Calculate all m values such that the roots of the following equation constitute an arithmetic sequence. x4 - (3m + 1) x2 + m2 = 0 (1)
Say r is a strictly positive root of the equation (1), then (-r) is a root too. Since the four roots form an arithmetic sequence, we can write these roots as -3r, -r, r, 3r.The only monic equation with these roots is
(x + 3r) (x + r) (x - r) (x - 3r) = 0 <=> x4 - 10 r2 x2 + 9 r4 = 0This equation is identical to (1) if and only if(3m + 1) = 10 r2 and 9 r4 = m2
- m = 3 r2
Then 9r2 + 1 = 10 r2 <=> ... <=> m = 3
- m = - 3 r2
Then - 9r2 + 1 = 10 r2 <=> ... <=> m = -3/19
- m = 3 r2
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In a sequence is the sum of the first n terms = sn = 2n.p - 1, with p = a fixed real number. Show that the sequence is geometric.
tn = sn+1 - sn = 2n.p + p - 1 - 2n.p + 1 = 2n.p + p - 2n.p = 2n.p.(2p - 1) tn-1 = 2n.p - p.(2p - 1) tn / tn-1 = 2p = independent of n = constant.
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