Question

prove [a+1/a]^2+[a-1/a]^2 = 2[a^2+1/a^2]^2​

Answer 1

Answer:Let n=1

Then  

a+(a+d)+(a+2d)+ ... +[a+(n-1)d] = a

On the other hand,

n/2[2a+(n-1)d] = 1/2 * 2a = a

So these expresions  

coincide.

Supose that we have proved the  

identity for all k

so  

in particular, for k=n we have that  

a+(a+d)+(a+2d)+ ... +[a+(n-1)d] = n/2[2a+(n-1)d]

We have to prove the identity for  

k=n+1, that  

is

a+(a+d)+(a+2d)+ ... +[a+(n-1)d] + [a+nd] = (n+1)/2 * [2a+nd]

Notice that by induction

a+(a+d)+(a+2d)+ ... +[a+(n-1)d] + [a+nd]  

= n/2[2a+(n-1)d] + [a+nd]  

=na + n(n-1) d/2 + a + nd  

=(n+1) a + (n^2/2 - n/2 + n) d  

=(n+1) a + (n^2/2 + n/2) d  

=(n+1) a + (n+1) n d /2

=(n+1)/2 [ 2 a + n d]  

Hence prove !

Step-by-step explanation:

Answer 2

$LHS = \Big( a + \frac{1}{a}\Big)^{2} + \Big( a - \frac{1}{a}\Big)^{2}$

$= a^{2} + 2 \times a \times \frac{1}{a} + \frac{1}{a^{2}} + a^{2} - 2 \times a \times \frac{1}{a} + \frac{1}{a^{2}}$

$= a^{2} + 2 + \frac{1}{a^{2}} + a^{2} - 2 + \frac{1}{a^{2}}$

$= a^{2} + \frac{1}{a^{2}} + a^{2} + \frac{1}{a^{2}}$

$= 2a^{2} + \frac{2}{a^{2}}$

$= 2\Big( a^{2} + \frac{1}{a^{2}}\Big)$

$= RHS$

$Hence\: proved$

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