indukcja matematyczna

[dargmagic]

Udowodnić, że dla każdej liczby naturalnej n
a)\(1 ^{2}+3 ^{2}+5 ^{2}+...+(2n-1) ^{2} = \frac{n}{3}(4n ^{2}-1)\)
b)\(1*2+2*3+3*4+...+n*(n+1)= \frac{n}{3}(n+1)(n+2)\)

[anka]

a)\(1 ^{2}+3 ^{2}+5 ^{2}+...+(2n-1) ^{2} = \frac{n}{3}(4n ^{2}-1)\)

1. n=1
Sprawdź sama
2. \(n=k\)
\(1 ^{2}+3 ^{2}+5 ^{2}+...+(2k-1) ^{2} = \frac{k}{3}(4k ^{2}-1)\)
3. \(n=k+1\)
\(1 ^{2}+3 ^{2}+5 ^{2}+...+(2k-1) ^{2} +[2(k+1)-1] ^{2}= \frac{k+1}{3}[4(k+1) ^{2}-1)]\)
\(1 ^{2}+3 ^{2}+5 ^{2}+...+(2k-1) ^{2} +(2k+1)^{2}= \frac{k+1}{3}(4k^2 + 8k + 3)\)
\(1 ^{2}+3 ^{2}+5 ^{2}+...+(2k-1) ^{2} +(2k+1)^{2}= \frac{k+1}{3}(2k + 1)(2k + 3)\)

\(L=1 ^{2}+3 ^{2}+5 ^{2}+...+(2k-1) ^{2} +(2k+1)^{2}=\frac{k}{3}(4k ^{2}-1)+(2k+1)^{2}=\frac{k}{3}(2k-1)(2k+1)+(2k+1)^{2}=\\
\frac{2k+1}{3}[k(2k-1)+3(2k+1)]=\frac{2k+1}{3}[2k^2-k+6k+3]=\frac{2k+1}{3}(2k^2+5k+3)=\frac{2k+1}{3}(k + 1)(2k + 3)=P\)


b)\(1*2+2*3+3*4+...+n*(n+1)= \frac{n}{3}(n+1)(n+2)\)

1. \(n=1\)
Sprawdź sama

2. \(n=k\)
\(1*2+2*3+3*4+...+k*(k+1)= \frac{k}{3}(k+1)(k+2)\)

3. \(n=k+1\)
\(1*2+2*3+3*4+...+k*(k+1)+(k+1)*(k+2)= \frac{k+1}{3}(k+2)(k+3)\)

\(L=1*2+2*3+3*4+...+k*(k+1)+(k+1)*(k+2)=\frac{k}{3}(k+1)(k+2)+(k+1)*(k+2)=\\
\frac{(k+1)(k+2)}{3}(k+3)=P\)