# Thread: Mathematical induction -P2 (priority 1)

1. ## Mathematical induction -P2 (priority 1)

Hi,

Can someone pls help urgently to solve the attached question using mathematical induction.

Lalit Chugh[B]

2. Hello, lalitchugh!

If you can't solve it, it's because there is an awful typo . . .

Prove the statement using mathematical induction:

. . . $1 + A + A^2 + A^3 +\,\cdots\,+ A^{n-1}\;=\;\frac{A^n - 1}{A - 1}$

Verify $S(1):\;\;1 \:=\:\frac{A^1 - 1}{A - 1} \;=\;1$ . . . True!

Assume $S(k):\;\;1 + A + A^2 + \hdots + A^{k-1}\;=\;\frac{A^k - 1}{A - 1}$

Add $A^k$ to both sides:
. . . $1 + A + A^2 + \hdots + A^{k-1} + A^k\;=\;\frac{A^k - 1}{A - 1} + A^k$

. . . $1 + A + A^2 + \hdots + A^k \;= \;\frac{A^k - 1}{A - 1} + \frac{A^k(A - 1)}{A - 1}$

. . . $1 + A + A^2 + \hdots + A^k \;=\;\frac{A^k - 1 + A^{k+1} - A^k}{A - 1}$

. . . $1 + A + A^2 + \hdots + A^k\;=\;\frac{A^{k+1} - 1}{A - 1}$ . . . This is $S(k+1)$ . . . We're done!

3. ## Mathematical induction - P2 (priority 1)

Hi,

Thanks for the solution But you have taken the R.H.S as $\frac{A^n - 1}{A - 1}$

And my question is like this:

Prove the statement using mathematical induction:

$1 + A + A^2 +.....+ A^n^-^1 = \frac{A^n^-^1}{A - 1}$

Request you to provide solution for this asap.

Thanks a lot
Lalit Chugh

4. Hello, Lalit!

I'll say it again: The statement is not true.

Prove the statement using mathematical induction:

$1 + A + A^2 + \hdots + A^{n-1} = \frac{A^{n-1}}{A - 1}$

Did you try any values of $$A$$ and $n$ ?

Let $A = 3$ and take 5 terms $(n = 5).$

. . $1 + 3 + 3^2 + 3^3 + 3^4 \;\;(?=?) \;\;\frac{3^4}{3 - 1}\;\begin{array}{ccc}\Leftarrow\\ \\ \end{array}$
. . . . . . . . . . . . . . $\Uparrow$

The left side is certainly greater than $3^4.$

The right side is certainly less than $3^4.$

. . Therefore . . .

5. $A\not = 1$

6. Originally Posted by lalitchugh
Hi,
Thanks for the solution But ...my question is like this:
...
$1 + A + A^2 +.....+ A^n^-^1 = \frac{A^n^-^1}{A - 1}$
...
Hello,

as Soroban pointed out you must have made a typo. Calculate the sum the more conventional way, you'll get n summands forming the sum s:

$\; \; \; S = 1 +A+A^2+A^3+...+A^{n-1}$. Now multiply this equation by A and you'll get:

$A\cdot S = \ \ \ A+A^2+A^3+...+A^{n-1}+A^{n}$

Now subtract:

$A\cdot S - S= A^{n}-1\ \Longrightarrow\ S(A-1)=A^{n}-1$

Divide by (A-1) and you'll get the formula given by Soroban.

Greetings

EB

7. Originally Posted by ThePerfectHacker
$A\not = 1$
HEY EVEN I THINK THE STATEMENT U GAVE IS WRONG CUZ IF YOU TRY IT FOR A=2 THEN ALSO IT DOESNT HOLD EQUALITY
ACCORDING TO ME A[HTML]<SUP>n</SUP>[/HTML] -1 / A-1 IS CORRECT
MAYBE YOU COPIED IT WRONG OR THE QUESTION WAS WRONG IN ITSELF

8. Originally Posted by killer baby
Originally Posted by ThePerfectHacker
$A\not = 1$
HEY EVEN I THINK THE STATEMENT U GAVE IS WRONG CUZ IF YOU TRY IT FOR A=2 THEN ALSO IT DOESNT HOLD EQUALITY
ACCORDING TO ME A[HTML]<SUP>n</SUP>[/HTML] -1 / A-1 IS CORRECT
MAYBE YOU COPIED IT WRONG OR THE QUESTION WAS WRONG IN ITSELF
Some rules that will help make your posts more intelligable:

1. Learn to quote the post you are responding to. Your comment is not on PH's post (which in itself is not easy to understand).

2. If you knoe it, use conventional English

3. All upper case (that's capital letters) is considered shouting, and in consequence rude.

RonL