From GMB (page 159), also proposed for 5th grade.
" E : 17461. Let $N=3+3^2+3^3+\dots+3^{4n+1}\;,\;n\in\mathbb{N}.$
a) Show that $2N+3$ s a perfect square.
b) For $n$ even natural numbers, determine the last two digits of $N$.
{author : } Marin CHIRCIU, Pitești "
Partial ANSWER CiP
a) $2N+3=(3^{2n+1})^2$
b) The answer is not a unique number. We have :
$3+3^2+3^3+3^4+3^5+3^6+3^7+3^8+3^9=29\;5\color{Yellow}{23}$
$3+3^2+3^3+\dots+3^{17}=193\;710\;2\color{Pink}{43}$
Solution CiP
a) $2N+3=\color{Red}{3}+N+N=$
$=\color{Red}{3}+(3+3)+(3^2+3^2)+(3^3+3^3)+\dots+(3^{4n+1}+3^{4n+1})=$
$=\color{Red}{(3+3+3)}+(3^2+3^2)+(3^3+3^3)+\dots+(3^{4n+1}+3^{4n+1}).$
Let's observe that $3+3+3=3^2$ ...and in general
$3^k+3^k+3^k=3^{k+1} \tag{1}$
Then, the red parenthesis results is $\color{Red}{3^2}$ , which appears as the term in the second parenthesis thus :
$(\color{Red}{3^2}+3^2+3^2)+(3^3+3^3)+\dots +(3^{4n+1}+3^{4n+1}).$
Continuing this calculation, we finally obtain the result
$2N+3=\color{Red}{3^{4n+1}}+3^{4n+1}+3^{4n+1}\underset{(1)}{=}3^{4n+2}=(3^{2n+1})^2\;,$
i.e. a perfect square.
$\square$
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