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Dr.Professor
 one year ago
Best ResponseYou've already chosen the best response.2Hi... Do you know what type of problem is that

aussy123
 one year ago
Best ResponseYou've already chosen the best response.1Yea its an induction problem.

Dr.Professor
 one year ago
Best ResponseYou've already chosen the best response.2Great.. I have a request for you... pls write the question here

aussy123
 one year ago
Best ResponseYou've already chosen the best response.1Ok Ill try my best to put it on here correctly

Dr.Professor
 one year ago
Best ResponseYou've already chosen the best response.2Thanks.. for that

aussy123
 one year ago
Best ResponseYou've already chosen the best response.1Ok Prove the statement by mathematical induction. 3 + 5 + 7 + . . . + (2n + 1) = n(n + 2) 1. proposition is true when n = 1, since n(n + 2) = 1(1 + 2) =3 2. We will assume that the proposition is true for a constant k = n so, 3 + 5 + 7 + . . . + (2k + 1) = __________(k + __________) 3. Then, 3 + 5 + 7 + . . . + (2k + 1) + (_____k + _____) = k(k + 2) + (________k + _______)

Dr.Professor
 one year ago
Best ResponseYou've already chosen the best response.2for the statement n= 1, the state ment reduces to\[1^2= \frac { 1\cdot 2\cdot 3 }{ 6 } \] and is obviously true. Assuming the statement is true for n = k: \[{ 1 }^{ 2 }+{ 2 }^{ 2 }+{ 3 }^{ 2 }+.....+{ 4 }^{ 2 }=\frac { k(k+1)(2k+1) }{ 6 } \] , we will prove that the statement must be true for n = k + 1: \[{ 1 }^{ 2 }+{ 2 }^{ 2 }+{ 3 }^{ 2 }+.....+{ (k+1) }^{ 2 }=\frac { (k+1)(k+2)(2k+3) }{ 6 } \] The lefthand side of (2) can be written as \[{ 1 }^{ 2 }+{ 2 }^{ 2 }+{ 3 }^{ 2 }+....+{ k }^{ 2 }+{ (k+1) }^{ 2 } \] In view of (1), this simplies to: \[{ (1 }^{ 2 }+{ 2 }^{ 2 }+{ 3 }^{ 2 }+....+{ k }^{ 2 })+{ (k+1) }^{ 2 }=\frac { k(k+1)(2k+1) }{ 6 } +{ (k+1) }^{ 2 }\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad \frac { k(k+1)(2k+1)+6{ (k+1) }^{ 2 } }{ 6 } \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad \frac { k(k+1)[k(2k+1)+6{ (k+1) }^{ 2 }] }{ 6 } \quad \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad \frac { (k+1)({ 2k }^{ 2 }+7k+6) }{ 6 } \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad \frac { (k + 1)(k + 2)(2k + 3) }{ 6 } \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \]

Dr.Professor
 one year ago
Best ResponseYou've already chosen the best response.2for the last part as its not clear

Dr.Professor
 one year ago
Best ResponseYou've already chosen the best response.2\[{ (1 }^{ 2 }+{ 2 }^{ 2 }+{ 3 }^{ 2 }+....+{ k }^{ 2 })+{ (k+1) }^{ 2 }=\frac { k(k+1)(2k+1) }{ 6 } +{ (k+1) }^{ 2 }\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad \frac { k(k+1)(2k+1)+6{ (k+1) }^{ 2 } }{ 6 } \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad \frac { k(k+1)[k(2k+1)+6{ (k+1) }^{ 2 }] }{ 6 } \quad \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad \frac { (k+1)({ 2k }^{ 2 }+7k+6) }{ 6 } \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad \frac { (k+1)(k+2)(2k+3) }{ 6 } \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \]

Dr.Professor
 one year ago
Best ResponseYou've already chosen the best response.2Now lets solve the second one

aussy123
 one year ago
Best ResponseYou've already chosen the best response.1I realy dont know what you did here, but My work didnt ask for all this. Its really confusing.

Dr.Professor
 one year ago
Best ResponseYou've already chosen the best response.2The solution for 2

Dr.Professor
 one year ago
Best ResponseYou've already chosen the best response.2or goto this http://home.cc.umanitoba.ca/~thomas/Courses/textS121.pdf
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