Question

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Solve this one step by step !

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Answer 1

The given sum is,

$\longrightarrow S=1+49+49^2+\,\dots\,+49^{125}$

which is a geometric series of $a=1,\ r=49>1$ and $n=126.$

Thus,

$\longrightarrow S=\dfrac{a(r^n-1)}{r-1}$

$\longrightarrow S=\dfrac{49^{126}-1}{49-1}$

Since $n^2-1=(n+1)(n-1),$

$\longrightarrow S=\dfrac{(49^{63}+1)(49^{63}-1)}{49-1}$

Since $\displaystyle\dfrac{a^n-1}{a-1}=\sum_{r=1}^na^{n-r},$

$\displaystyle\longrightarrow S=(49^{63}+1)\sum_{r=1}^{63}49^{63-r}$

$\longrightarrow S=(49^{63}+1)(49^{62}+49^{61}+49^{60}+\,\dots\,+1)$

This implies,

$\longrightarrow (49^{63}+1)\ \Big|\ S$

And therefore,

$\longrightarrow\underline{\underline{k=63}}$

Hence (D) is the answer.

Shortcut:-

We see that,

$\begin{aligned}\longrightarrow\ \ &(1+a+a^2+\,\dots\,+a^{k-1})(1+a^k+a^{2k}+\,\dots\,+a^{nk})\\=\ \ &1+a+a^2+\,\dots\,+a^{(n+1)k-1}=S\quad\quad\dots(2)\end{aligned}$

This implies,

$\longrightarrow (1+a^k+a^{2k}+\,\dots\,+a^{nk})\ \Big|\ S$

which can be reduced as,

$\longrightarrow (1+a^k)\ \Big|\ S$

if and only if $n=1.$

Then (2) becomes,

$\longrightarrow(1+a+a^2+\,\dots\,+a^{k-1})(1+a^k)=1+a+a^2+\,\dots\,+a^{2k-1}=S\quad\quad\dots(3)$

Let,

$\longrightarrow S=1+a+a^2+\,\dots\,+a^m\quad\quad\dots(4)$

Comparing (3) and (4),

$\longrightarrow m=2k-1$

$\longrightarrow\boxed{k=\dfrac{m+1}{2}}$

Therefore,

$\boxed{\begin{minipage}{9.7cm}The greatest positive integer k for which (a^k+1) is a factor of\\ 1+a+a^2+\,\dots\,+a^m is,\begin{center} k=\dfrac{m+1}{2} \end{center}\end{minipage}}$

In the question, $m=125.$

So the answer is,

$\longrightarrow k=\dfrac{125+1}{2}$

$\longrightarrow\underline{\underline{k=63}}$

Hence (D) is the answer.

Answer 2

Answer:

The answer of this question is option d.