Točno
21. listopada 2013. 16:06 (10 godine, 9 mjeseci)
Sakrij rješenje
Dokaži indukcijom da je suma prvih
![n](/media/m/a/e/5/ae594d7d1e46f4b979494cf8a815232b.png)
kvadrata prirodnih brojeva jednaka
![\frac{n(n+1)(2n+1)}{6}](/media/m/a/5/d/a5de3f1eaaad4e34e5a5ca7c7be7c291.png)
.
%V0
Dokaži indukcijom da je suma prvih $n$ kvadrata prirodnih brojeva jednaka $\frac{n(n+1)(2n+1)}{6}$.
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Baza![n=1](/media/m/4/e/4/4e466fe58c2a8f6389234c5c673f069c.png)
![1^2=\dfrac{1(1+1)(2+1)}{6}](/media/m/e/8/3/e83803518ce161a736c7697e2edc3455.png)
![1=\dfrac{1\cdot 2 \cdot 3}{6}](/media/m/5/0/2/5028f6c33616b746105f60416d428159.png)
![1=1](/media/m/8/1/a/81a2503d9863ddb873ecfffeb0c86ac1.png)
Ova tvrdnja ocito vrijedi
PretpostavkaPretpostavimo da vrijedi
![1^2+2^2+...+n^2=\dfrac{n(n+1)(2n+1)}{6}](/media/m/5/8/0/5804585f680a97ec477f9871fd1fd9c1.png)
za neki prirodan broj
KorakZelimo dokazati da
![1^2+2^2+...+n^2+(n+1)^2=\dfrac{(n+1)(n+2)(2(n+1)+1)}{6}](/media/m/3/5/8/3586ea81424ca0ea9cacda55a6186caf.png)
![1^2+2^2+...+n^2+(n+1)^2=\dfrac{(n+1)(n+2)(2n+3)}{6}](/media/m/5/a/d/5adc85fd00d91b85cddf7a273319a386.png)
Iz pretpostavke znamo da je
![1^2+2^2+...+n^2=\dfrac{n(n+1)(2n+1)}{6}](/media/m/5/8/0/5804585f680a97ec477f9871fd1fd9c1.png)
, pa uvrstimo to
Dobivamo:
![\dfrac{n(n+1)(2n+1)}{6}+(n+1)^2=\dfrac{(n+1)(n+2)(2n+3)}{6}](/media/m/1/6/b/16b6579763a212d1acfd7dd962b9b299.png)
![\dfrac{n(n+1)(2n+1)+6(n+1)^2}{6}=\dfrac{(n+1)(n+2)(2n+3)}{6}](/media/m/7/0/c/70c708b727d57e09201aa76fb7d03da0.png)
Djeljenjem cijele jednadzbe sa
![n+1](/media/m/2/a/7/2a7327e09a84d01a602088c9f045cbde.png)
i mnozenjem sa
![6](/media/m/e/e/e/eeec330d59a70f8ed1d6882474cb02a3.png)
dobivamo:
![n(2n+1) + 6(n+1)=(n+2)(2n+3)](/media/m/1/3/1/131910e2891dfd91300e25ea8ef9268b.png)
Sto ocito vrijedi
%V0
[b]Baza[/b]
$n=1$
$1^2=\dfrac{1(1+1)(2+1)}{6}$
$1=\dfrac{1\cdot 2 \cdot 3}{6}$
$1=1$
Ova tvrdnja ocito vrijedi
[b]Pretpostavka[/b]
Pretpostavimo da vrijedi
$1^2+2^2+...+n^2=\dfrac{n(n+1)(2n+1)}{6}$
za neki prirodan broj $n$
[b]Korak[/b]
Zelimo dokazati da
$1^2+2^2+...+n^2+(n+1)^2=\dfrac{(n+1)(n+2)(2(n+1)+1)}{6}$
$1^2+2^2+...+n^2+(n+1)^2=\dfrac{(n+1)(n+2)(2n+3)}{6}$
Iz pretpostavke znamo da je
$1^2+2^2+...+n^2=\dfrac{n(n+1)(2n+1)}{6}$, pa uvrstimo to
Dobivamo:
$\dfrac{n(n+1)(2n+1)}{6}+(n+1)^2=\dfrac{(n+1)(n+2)(2n+3)}{6}$
$\dfrac{n(n+1)(2n+1)+6(n+1)^2}{6}=\dfrac{(n+1)(n+2)(2n+3)}{6}$
Djeljenjem cijele jednadzbe sa $n+1$ i mnozenjem sa $6$ dobivamo:
$n(2n+1) + 6(n+1)=(n+2)(2n+3)$
$2n^2 +7n+6=2n^2 + 7n +6$
Sto ocito vrijedi
21. listopada 2013. 19:29 | ikicic | Točno |