Let \(\displaystyle *^{n}(a) = {a*.......*a } \) n times
thus
\(\displaystyle *^{1} = a \)
\(\displaystyle *^{2} = a*a \)
\(\displaystyle *^{3} = a*a*a \) etc
prove by induction that
\(\displaystyle *^{n}a = (a+1)^{n} -1 \) for all n in N
am really having problems understand how to do this, please can someone show a step by step solution guide?
so far i get
n = 1
\(\displaystyle *^{1}a = a \)
\(\displaystyle (a+1)^{1} -1 = a \)
so true for n = 1
assume true for n = k
so
\(\displaystyle *^{k}a = (a+1)^{k} - 1 \)
next step in indution, cant seem to do it.....
n= k+1
\(\displaystyle *^{k+1}a = *^{k}a *a \)
thats equal to
\(\displaystyle (a+1)^{k}-1*a \)
where do i go from here ?
thus
\(\displaystyle *^{1} = a \)
\(\displaystyle *^{2} = a*a \)
\(\displaystyle *^{3} = a*a*a \) etc
prove by induction that
\(\displaystyle *^{n}a = (a+1)^{n} -1 \) for all n in N
am really having problems understand how to do this, please can someone show a step by step solution guide?
so far i get
n = 1
\(\displaystyle *^{1}a = a \)
\(\displaystyle (a+1)^{1} -1 = a \)
so true for n = 1
assume true for n = k
so
\(\displaystyle *^{k}a = (a+1)^{k} - 1 \)
next step in indution, cant seem to do it.....
n= k+1
\(\displaystyle *^{k+1}a = *^{k}a *a \)
thats equal to
\(\displaystyle (a+1)^{k}-1*a \)
where do i go from here ?