Question

Consider the expression 11^n.

Answer 1

Answer:

A. 11^0=1, 11^1=11, 11^2=121, 11^3=1331, 11^4=16461

B. if n is an integer, the digits of the results are the same as the coefficient in the n-row of pascal´s triangle.

C. 11^5=161051 and, no, it does not match the pattern

Step-by-step explanation:

A. Used a calculator 11^0=1, 11^1=11, 11^2=121, 11^3=1331, 11^4=16461

B. Visual, if n is an integer, the digits of the results are the same as the coefficient in the n-row of pascal´s triangle.

C. it´s useful to rewrite 11 as 10+1, now we are able to use the binomial theorem:

$(10+1)^{n}=\sum_{i=0}^{n}{{n\choose{i}}(10^{i})(1^{n-i})$

let´s test this with n=2 and see what happens:

$(10+1)^{2}= {2\choose{0}}(1)+{2\choose{1}}(10) +{2\choose{2}}(100)$

What we observed, is that "10^i" places the combinatoric number in digits based in the value of i (i=0 first digit, i=1 second digit ans so on)

(10+1)^2=1+2(10)+1(100)

11^2=121

Notice the combinatoric numbers are less than 10, so they can be placed into those places as digits. This would work until the combinatoric numbers are greater or equal to 10.

To prove it we need the hypothesis that n≤4 (we assure the combinatoric numbers are less than 10), now we use the binomial theorem.

$(10+1)^{n}=\sum_{i=0}^{n}{{n\choose{i}}(10^{i})(1^{n-i})$

$(10+1)^{n}= {n\choose{0}}(1)+{n\choose{1}}(10)+{n\choose{2}}(100)+{n\choose{3}}(1000)+{n\choose{4}}(10000))$

Considering that the combinatoric numbers are less than 10 (they are single digits) and knowing the term "10^i" places the combinatoric number in digits (i=0 first digit, i=1 second digit ans so on), we can affirm:

$(10+1)^{n}= "{n\choose{4}}"{n\choose{3}}"{n\choose{2}}"{n\choose{1}}"{n\choose{0}}"$

with the combinatoric numbers as digits. And we proved what we were asked for

D. 11^5=161051, and it does not follow the pattern. That´s because ${5\choose{2}}=10$ and we have shown that it only work with the combinatoric numbers being single digits. Because 10 occupies 2 digits places and multiplying it by "10^i" will interfere with the other digits placements.