Ejercicio 4. Demostrar, para $a$, $d$ fijos y $n$ variable
$a+(a+d)+(a+2d)+\cdots+\left(a+(n-1)d \right)= \frac{n}{2}\left(2a+(n-1)d \right).$
Demostración:
Para: $n=1$, $a+(1-1)d= \frac{1}{2}\left(2a+(1-1)d \right)$
$a=a$. $\checkmark$
$n=k$,
HI $a+(a+d)+(a+2d)+\cdots+\left(a+(k-1)d \right)= \frac{k}{2}\left(2a+(k-1)d \right).$ $\checkmark$
$n=k^*$,
TI $a+(a+d)+(a+2d)+\cdots+\left(a+(k^*-1)d \right)=\underbrace{a+(a+d)+(a+2d)+\cdots+\left(a+(k-1)d \right)}+\left(a+(k^*-1)d \right)$ HIDemostración:
Para: $n=1$, $a+(1-1)d= \frac{1}{2}\left(2a+(1-1)d \right)$
$a=a$. $\checkmark$
$n=k$,
HI $a+(a+d)+(a+2d)+\cdots+\left(a+(k-1)d \right)= \frac{k}{2}\left(2a+(k-1)d \right).$ $\checkmark$
$n=k^*$,
$=\underbrace{\frac{k}{2}\left(2a+(k-1)d \right)}+\left(a+(k^*-1)d \right)$ HI
$=\frac{k\left(2a+(k-1)d \right)+2\left(a+(k^*-1)d \right)}{2}$
$=\frac{k\left(2a+(k-1)d \right)+2\left(a+\left((k+1)-1\right)d \right)}{2}$
$=\frac{k\left(2a+kd-d \right)+2\left(a+kd \right)}{2}$
$=\frac{k2a+k^2d-kd +2a+2kd }{2}$
$=\frac{k2a+2a+k^2d+kd }{2}$
$=\frac{\left(k+1\right)2a+kd\left(k+1\right) }{2}$
$=\frac{\left(k+1\right)\left(2a+kd\right) }{2}$
$=\frac{\left(k+1\right)\left[2a+\left(\left(k+1\right)-1\right)d\right] }{2}$
$=\frac{k^* \left[2a+\left(k^*-1\right)\right]d}{2}$ $\checkmark$
Ejercicio 5. Demostrar $2+2^2+2^3+\cdots2^n=2\left(2^n-1\right).$
Para: $n=1$, $2^1=2\left(2^1-1\right)=2\left(2-1\right)=2\cdot 1=2.$ $\checkmark$
HI $n=k$, $2+2^2+2^3+\cdots2^k=2\left(2^k-1\right).$ $\checkmark$
TI $n=k^*$ $2+2^2+2^3+\cdots2^{k^*}=2\left(2^{k^*}-1\right).$
Prueba: $2+2^2+2^3+\cdots2^{k^*}=\underbrace{2+2^2+2^3+\cdots2^k}+2^{k^*}$
HI
$=\underbrace{2\left(2^k-1\right)}+2^{k^*}$
HI
$=2^{k+1}-2+2^{k+1}$
$=2\cdot 2^{k+1}-2$
$=2\left( 2^{k^*}-1\right).$ $\checkmark$
Ejercicio 6. Demostrar:
$1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}\cdots+\frac{1}{2^{n-1}}=2-\frac{1}{2^{n-1}}.$
Para: $n=1$, $\frac{1}{2^{1-1}}= 2-\frac{1}{2^{1-1}}$
$\frac{1}{2^0}= 2-\frac{1}{2^0}$
$\frac{1}{1}= 2-\frac{1}{1}$
$1=1.$ $\checkmark$
$n=k$
HI
$1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}\cdots+\frac{1}{2^{k-1}}=2-\frac{1}{2^{k-1}}.$
$\checkmark$
$n=k^*$
TI
$1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}\cdots+\frac{1}{2^{k^*-1}}=2-\frac{1}{2^{k^*-1}}.$
Prueba:
$1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}\cdots+\frac{1}{2^{k^*-1}}=\underbrace{1+\frac{1} {2}+\frac{1}{2^2}+\frac{1}{2^3}\cdots+\frac{1}{2^{k-1}}}+\frac{1}{2^{k^*-1}}$
$=\underbrace{2-\frac{1}{2^{k-1}}} +\frac{1}{2^{k^*-1}}$ HI
$=2-\frac{1}{2^k\cdot 2^{-1}} +\frac{1}{2^{\left(k+1\right)-1}}$
$=2-\frac{2}{2^k} +\frac{1}{2^k}$
$=2-\frac{1}{2^k} =2-\frac{1}{2^{\left(k+1\right)-1}}$
$=2-\frac{1}{2^{k^*-1}}$ $\checkmark$
Ejercicio 7. Demostrar
$1^3+2^3+3^3+\cdots+n^3=\frac{n^2}{4}\left(n+1\right)^2.$
Demostración:
Para: $n=1$, $1^3=\frac{1^2}{4}\left(1+1\right)^2$
$1=\frac{1^2}{4}2^2$
$1=\frac{1}{4}4$
$1=1$ $\checkmark$
HI
Para $n=k$,
$1^3+2^3+3^3+\cdots+k^3=\frac{k^2}{4}\left(k+1\right)^2.$ $\checkmark$
TI
Para $n=k^*$,
$1^3+2^3+3^3+\cdots+\left(k^*\right)^3=\frac{\left(k^*\right)^2}{4}\left(k^*+1\right)^2.$
Demostración:
$1^3+2^3+3^3+\cdots+\left(k^*\right)^3=\underbrace{1^3+2^3+3^3+\cdots+k^3}+\left(k^*\right)^3$
$=\underbrace{\frac{k^2}{4}\left(k+1\right)^2}+\left(k+1\right)^3$
$=\left(k+1\right)^2\left(\frac{k^2}{4}+k+1\right)$
$=\left(k+1\right)^2 \frac{\left(k^2+4k+4\right)}{4}$
$=\frac{\left(k+1\right)^2}{4}\left(k+2\right)^2$
$=\frac{\left(k+1\right)^2}{4}\left(\left(k+1\right)+1\right)$
$=\frac{\left(k^*\right)^2}{4}\left(k^*+1\right)^2.$ $\checkmark$
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