Proof by Induction
Proof by induction basically proves #P(n)# for all integers #n≥b# by proving that #P(c)# is true and that #P(k+1)# is true whenever #P(k)# is true.
So, the question asks us to prove that #3*2^2+3^2*2^3+…+3^n*2^(n+1)=12/5*(6^n-1)#, or, using sigma notation,
#sum_(i=1)^n3^i2^(i+1)=12/5(6^n-1)#
First, we define #P(k)# as true if the above sum is true when #n=k#. Now, according to the method of induction, we first show that it is true for some base case #b#, which the method of induction will then show that #P(n)# for all #n≥b#.
Here, we choose #c=1# and demonstrate #P(1)#, as
#sum_(i=1)^(1)3^i2^(i+1)=12=12/5(6^1-1)#
Then, the second step is to prove that #P(k+1)# is true whenever #P(k)# is true. In other words, if we assume that
#sum_(i=1)^k3^i2^(i+1)=12/5(6^k-1)#
is true, then
#sum_(i=1)^(k+1)3^i2^(i+1)=12/5(6^(k+1)-1)#
is true.
So, we start with
#sum_(i=1)^(k+1)3^i2^(i+1)#
which, pulling out the term for #i=k+1#, becomes
#=3^(k+1)2^(k+2)+sum_(i=1)^k3^i2^(i+1)#
Now, since we assumed that #sum_(i=1)^k3^i2^(i+1)=12/5(6^k-1)#, the above equals
#=3^(k+1)2^(k+2)+12/5(6^k-1)#
#=12/5*5*3^k2^k+12/5(6^k-1)#
#=12/5(5*3^k2^k+6^k-1)#
#=12/5(5*6^k+6^k-1)#
#=12/5(6*6^k-1)#
#=12/5(6^(k+1)-1)#
Thus, we have proven that #P(1)# is true, and #P(k)# is true whenever #P(k+1)# is true. Therefore, #P(n)# is true for all integers #n≥1#. (In other words, since #P(1)# is true, #P(2)# is true, #P(3)# is true, and so on.)
So,
#3*2^2+3^2*2^3+…+3^n*2^(n+1)=sum_(i=1)^n3^i2^(i+1)=12/5(6^n-1)#
#"Q.E.D."#
