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

1. aussy123

2. aussy123

In attachment

3. Dr.Professor

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

4. aussy123

Yea its an induction problem.

5. Dr.Professor

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

6. aussy123

Ok Ill try my best to put it on here correctly

7. Dr.Professor

Thanks.. for that

8. aussy123

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 + _______)

9. Dr.Professor

so 1st one,

10. Dr.Professor

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

11. Dr.Professor

for the last part as its not clear

12. Dr.Professor

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

13. Dr.Professor

Now lets solve the second one

14. aussy123

okay

15. aussy123

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

16. Dr.Professor

The solution for 2

17. Dr.Professor
18. aussy123

ok Ill try this..