## Mathsadness 5 years ago sum{k=1 to 30} (6k^2+1) How would you find the value of this sum?

1. anonymous

sum as $6\sum_1^{30}k^2 +\sum_1^{30}1$

2. anonymous

second term is just 30

3. anonymous

do you know the formula for $\sum_{k=1}^n k^2$?

no

5. anonymous

its is $\frac{n(n+1)(2n+1)}{6}$

6. anonymous

replace n by 30, compute and you will have the answer. i will write it if you like

7. anonymous

$6\times \frac{30(31)(71)}{6} +30$ $30(31)(71)+30$

8. anonymous

i get 66060

$\sum_{k=1}^{n} k^3$ What is the formula for this??

...

11. anonymous

that one is easy to remember.

12. anonymous

do you know the one for $\sum_{k=1}^{n}k$?

13. anonymous

it is $\frac{n(n+1)}{2}$

14. anonymous

the one for $\sum_{k=1}^n k^3$ is the square of that one. it is$(\frac{n(n+1)}{2})^2$

so the square of Gauss' formula??

16. anonymous

if that thing is called gauss's formula then yes

for the summation of n positive integers....

18. anonymous

yes. here they are, more than you want to know http://polysum.tripod.com/

wow!

20. anonymous

oh yes, i see that it is called gauss' formula. i learn more here than in skule

21. anonymous

if you like i can show you how to derive them, but you probably don't need to know it

how do you do that?

:P

24. anonymous

ok we start with the first one. "gauss" and the method extends. you want $\sum_{k=1}^n k$

25. anonymous

write this $\sum_{k=1}^n k^2-(k-1)^2$

26. anonymous

it might not be obvious, but the part after the sigma is one term minus the previous one. so this is a telescoping sum, and when you are done (it will be obvious if you write out some terms) everything will be killed off except the $n^2$

27. anonymous

so by observation we have $\sum_{k=1}^n k^2-(k-1)^2=n^2$

28. anonymous

then do the algebra: $k^2-(k-1)^2 = k^2-(k^2-2k+1)=2k-1$

29. anonymous

in other words there was no square in it. now we know that $\sum_{k=1}^n 2k-1=n^2$ because the part after the sum was the same. this is equal to $2\sum_{k=1}^nk-\sum_{k=1}^n =n^2$

30. anonymous

now solve for $\sum_{k=1}^nk$ using algebra and you get $\sum_{k=1}^n k = \frac{n(n+1)}{2}$

31. anonymous

f course you have probably seen this proved a different way. but now the fun begins. if you want the formal for $\sum_{k=1}^n k^2$ you write $\sum_{k=1}^n k^3-(k-1)^3=n^3$

32. anonymous

do the algebra and the $k^3$ will drop ouit (just like the k^2 did before) and you will get a summation only involving $k^2$ and k and 1. you already have the one for k, so solve for $\sum_{k=1}^nk^2$ and you will get your answer. this trick works all the way up, but the algebra gets really really annoying so best just to have a list

cool!

34. anonymous

problem 9 - 12 and and 20 - 23 work it out (not the algebra)

35. anonymous

have fun