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aussy123
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Can someone help me finish this problem? Medal Rewarded!
 one year ago
 one year ago
aussy123 Group Title
Can someone help me finish this problem? Medal Rewarded!
 one year ago
 one year ago

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aussy123 Group TitleBest ResponseYou've already chosen the best response.1
In attachment
 one year ago

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

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

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

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

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

aussy123 Group TitleBest ResponseYou've already chosen the best response.1
Ok 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 + _______)
 one year ago

Dr.Professor Group TitleBest ResponseYou've already chosen the best response.2
so 1st one,
 one year ago

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

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

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

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

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

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

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

aussy123 Group TitleBest ResponseYou've already chosen the best response.1
ok Ill try this..
 one year ago
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