Question

1. Let n E N, n # 0: (a) Show that n+1)* < m(n+1}* marks 1 marks 6 marks (b) Show that n+1) 1 n1 n (c) Use (a) and (b) to show by induction that < 2 - 1 Remark: This shows in particular that is finite, if you are interested the actual value of this sum is. See https://en.wikipedia.org/wiki/Basel problem for more info marks [8

I understand a) and b), but I do not know how to answer c). Can someone clearly explain and answer the last question, please!

Uploaded the previous question if that helps answer question 2!

Answer 1

1 + 5 o =

$1-\phi=\frac{1-\sqrt5}{2}$

From (a),

I + 0 = 20

From (b),

ي - 2 = 2( - 1)

Also,
V5-1
and
= ي - 1

And so,

) - انا = (1) و

let function f is true for some k and k-1 then-

(A) - to - -

now for k-1,

$f(k-1)=\frac{1}{\sqrt5} [\phi^{k-1}-(-\frac{1}{\phi})^{k-1}]$

Now

f(k+1) = f(k)f(k-1)

$f(k+1)=\frac{1}{\sqrt5} [\phi^{k}-(-\frac{1}{\phi})^{k}]+\frac{1}{\sqrt5} [\phi^{k-1}-(-\frac{1}{\phi})^{k-1}]$

5(x + 1) = klo1 +5-1-3--) f(k+1) =

We already have shown,
= ي - 1
, from this $1+\frac{1}{\phi}=\phi$

So we get, $f(k+1)=\frac{1}{\sqrt5} [\phi^{k+1}-(-\frac{1}{\phi})^{k+1}]$

Means, if formula is true for n=k-1 and n=k then it must be true for n=k+1

Now, for n=0,

f(0) = 0

for n=1 , $f(1)=1$

$f(2)=f(0)+f(1)=1$

$f(2)=\frac{1}{\sqrt5}[(\frac{1+\sqrt5}{2})^{2}-(\frac{1-\sqrt5}{2})^{2}]=1$

Hence, proved.

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