Question

Can someone help me finish this problem? Medal Rewarded!

Answer 1

In attachment

Answer 2

Hi...
Do you know what type of problem is that

Answer 3

Yea its an induction problem.

Answer 4

Great.. I have a request for you... pls write the question here

Answer 5

Ok Ill try my best to put it on here correctly

Answer 6

Thanks.. for that

Answer 7

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 + _______)

Answer 8

so 1st one,

Answer 9

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 left-hand 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:

\[{ (1 }^{ 2 }+{ 2 }^{ 2 }+{ 3 }^{ 2 }+....+{ k }^{ 2 })+{ (k+1) }^{ 2 }=\frac { k(k+1)(2k+1) }{ 6 } +{ (k+1) }^{ 2 }\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad \frac { k(k+1)(2k+1)+6{ (k+1) }^{ 2 } }{ 6 } \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad \frac { k(k+1)[k(2k+1)+6{ (k+1) }^{ 2 }] }{ 6 } \quad \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad \frac { (k+1)({ 2k }^{ 2 }+7k+6) }{ 6 } \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad \frac { (k + 1)(k + 2)(2k + 3) }{ 6 } \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \]

Answer 10

for the last part as its not clear

Answer 11

\[{ (1 }^{ 2 }+{ 2 }^{ 2 }+{ 3 }^{ 2 }+....+{ k }^{ 2 })+{ (k+1) }^{ 2 }=\frac { k(k+1)(2k+1) }{ 6 } +{ (k+1) }^{ 2 }\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad \frac { k(k+1)(2k+1)+6{ (k+1) }^{ 2 } }{ 6 } \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad \frac { k(k+1)[k(2k+1)+6{ (k+1) }^{ 2 }] }{ 6 } \quad \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad \frac { (k+1)({ 2k }^{ 2 }+7k+6) }{ 6 } \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad \frac { (k+1)(k+2)(2k+3) }{ 6 } \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \]

Answer 12

Now lets solve the second one

Answer 13

okay

Answer 14

I realy dont know what you did here, but My work didnt ask for all this. Its really confusing.

Answer 15

The solution for 2

Answer 16

or goto this http://home.cc.umanitoba.ca/~thomas/Courses/textS1-21.pdf

Answer 17

ok Ill try this..