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

1. aussy123 Group Title

2. aussy123 Group Title

In attachment

3. Dr.Professor Group Title

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

4. aussy123 Group Title

Yea its an induction problem.

5. Dr.Professor Group Title

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

6. aussy123 Group Title

Ok Ill try my best to put it on here correctly

7. Dr.Professor Group Title

Thanks.. for that

8. aussy123 Group Title

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 Group Title

so 1st one,

10. Dr.Professor Group Title

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 Group Title

for the last part as its not clear

12. Dr.Professor Group Title

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

13. Dr.Professor Group Title

Now lets solve the second one

14. aussy123 Group Title

okay

15. aussy123 Group Title

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

16. Dr.Professor Group Title

The solution for 2

17. Dr.Professor Group Title
18. aussy123 Group Title

ok Ill try this..