Adding Powers of 9

$\begin{aligned} 9^n + 9^{n+1} &= (9^n \times 1) + (9^n \times 9) \\ &= 9^n \times (1+9) \\ &= 9^n \times 10 \end{aligned}$

$9^{n}+9^{n+1}=9^{n}+9^{n}\times 9=9^{n}\times(1+9)=9^{n}\times 10$

For any positive integer $n$ :

• $9^n$ ends with $9$ if $n$ is an odd integer

• $9^n$ ends with $1$ if $n$ is an even number.

Now, take any two consecutive integers. In that two definitely one is an even number and one is an odd number.

So, here we will take $9^n$ and $9^{n + 1}$ . If $9^n$ is odd then $9^{n + 1}$ is even. So the last digit of first one will be $1$ and the last digit second one will be $9$ . And there sum will give $1 + 9 = 10$ . If $9^n$ is even then $9^{n + 1}$ will be odd and the last digits will get interchanged. So there sum will be $9 + 1 = 10$ .

So, in any case $9^n + 9^{n + 1}$ will be a multiple of $10$ as it will end up with digit $0$ .

$9^n+9^{n+1}=9^n\times 10$ $\frac{9^n}{9^n}+\frac{9^{n+1}}{9^n}=\frac{9^n\times 10}{9^n}$ $1+9=10$

Therefore, this statement is always true for any positive integer $n$ .

Using mathematical induction,

Let P(n) be the case that 9^n + 9^(n+1) = 9^n x 10 , n is positive integer

when n = 1, as shown in the question, is true

Therefore the base case is true

Assume true for n=k, i.e. 9^k + 9^(k+1) = 9^k x 10 -- (1) , k is positive integer

To prove true for n=k+1, i.e. need to show that 9^(k+1) + 9^(k+2) = 9^(k+1) x 10

LHS = 9^(k+1) + 9^(k+2) = 9(9^k) + 9(9^(k+1)) = 9(9^k + 9^(k+1))

Substituting (1) into LHS,

LHS = 9(9^k x 10) = 9^(k+1) x 10 = RHS

Hence, the (k+1)th case in true.

Therefore since P(0) is true, and P(k+1) is true => P(k) is true, by Mathematical Induction, the above equation is true for any positive integer n

$9^n+9^{n+1}=9^n(1+9)=\boxed{9^n \times 10}$

For any x, it is true that x + 9x = 10x. If x = 9^n, then 9x = 9^n * 9, or 9^{n+1}. This means that the original equation is in the form of “x + 9x” making the result 10x.

Divide both sides by $9^n$ and you will see that this leads to a correct statement.

$9^n + 9^{n+1} = 9^n* 10 \Longleftrightarrow 1 + 9 = 10$

$9^n+9^{n+1}=1 \times 9^n+9 \times 9^n=9^n(1+9)=10 \times 9^n$

$9^{n}+9^{n+1} = 9^{n} \times 10$ $\frac{9^{n}+9^{n+1} = 9^{n} \times 10}{9^n}$ $1+9^{n+1-n}=10$ $\boxed{1+9=10}$

The expression is indexed by a natural number n, thus we can use mathematical induction to show the result holds for all natural numbers.

First, let P(n) denote the statement given.

Basis Step: P(0) LHS: 9^0 + 9^1 = 10 RHS: (9^0) x 10 = 10 LHS = RHS thus P(n) holds true for n = 0.

Inductive Step: P(K), P(K + 1) Assume the statement holds true for some natural number K, that is,

9^K + 9^(K + 1) = (9^K) x 10. (I)

Now, we want to show the statement P(K +1) holds, that is,

9^[(K) +1] + 9^[(K + 1) + 1] = (9^[(K) + 1]) x 10.

If we multiply equation (I) by 9, we get

9.(9^K) + 9.[9^(K + 1)] = 9.(9^K) x 10

Now, if we utilise laws of indices from elementary algebra in secondary school

“(a^b).(a^c) = (a^[b + c]) for all real numbers a, b and c”

we obtain

9^[(K) +1] + 9^[(K + 1) + 1] = (9^[(K) + 1]) x 10.

Conclusion: Thus, we have shown for all natural numbers K and K + 1 (including zero in this instance) that the statement P(n) holds true. And so, it has been proven by mathematical induction.

By the Induction Principle:

We probe $9^n + 9^{n+1} = 9^n * 10$ for n = 0:

$9^0 + 9 ^{0+1} = 9^0 * 10 \implies 1 + 9 = 10$

Then we assume (1) $9^n + 9^{n+1} = 9^n * 10$ true and probe for (n+1): $9^{n+1} + 9^{n+2} = 9^{n+1} * 10$

$9^{n+1} + 9^{n+2} = 9*9^n + 9*9^{n+1} = 9 * (9^n + 9^{n+1})\implies by (1) = 9* (9^n * 10) = 9 * 9^n * 10 = 9^{n+1} * 10$

And QED.

If you like mods then here's the solution...

9 is congruent to -1 mod 10... Consider to cases... i) n is even ii) n is odd

i)If n is even... then 9^n is congruent to (-1)^an even number which is 1 If n is even.. it implies n+1 is odd.. which implies 9^n+1 is congruent to (-1)^an odd number which is -1 Hence 9^n + 9^n+1 = 1-1 = o mod 10

ii) Through similar arguments we find that 9^n + 9^n+1 = -1 + 1 = 0 mod 10

Therefore 9^n + 9^n+1 is always divisible by 10

10 = 9 + 1

Substituting this into the equation provided yields:

9^n + 9^{n + 1} = 9^n \times (9 + 1) 9^n + 9^{n + 1} = 9^{n + 1} + 9^n

Apologies, the markdown syntax doesn't seem to work on mobile.

Theoretically we can prove it by principle of mathematical induction.