SOLUTION: Hello, could I please have some help with this question, Prove by mathematical induction that; (1 × 2) + (3 × 4) + · · · + (2n − 1) × 2n = n(n + 1)(4n − 1)/3 for

Algebra ->  Sequences-and-series -> SOLUTION: Hello, could I please have some help with this question, Prove by mathematical induction that; (1 × 2) + (3 × 4) + · · · + (2n − 1) × 2n = n(n + 1)(4n − 1)/3 for      Log On


   

Question 1075196: Hello,
could I please have some help with this question,
Prove by mathematical induction that;
(1 × 2) + (3 × 4) + · · · + (2n − 1) × 2n = n(n + 1)(4n − 1)/3
for all n ∈ N.
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
The left hand side of the equation is the sum of
n products of the form %282n-1%29%2A2n ,
starting from %282%2A1-1%29%2A%282%2A1%29=%281%2A2%29 .
You need to prove that
1) it is true for n=1 , the fisrt natural number in N.
2) If it is true for n=k , it must be true for n=k%2B1 .

1) For n=1 the equation becomes
The equation becomes
%281%2A2%29=1%281%2B1%29%284%2A1-1%29%2F3 <---> 1%2A2=1%2A2%2A3%2F3 ,
which is obviously true.

2) For n=k the equation becomes
.
The sum on the left hand sides has k terms,
If the formula you want to prove is true for n=k ,
those k terms add to k%28k%2B1%29%284k-1%29%2F3 .

For n=k%2B1 the equation becomes

and that is what you have to prove, staring from
%281%2A2%29+%2B+%283%2A4%29+%2B+%22...%22+%2B+%282k-1%29+%2A+2k=k%28k%2B1%29%284k-1%29%2F3 .
The sum on the left hand sides have the same first k terms
up to %282k-1%292k , the term number k ,
%282%28k%2B1%29-1%29+%2A+2%28k%2B1%29

Starting from
%281%2A2%29+%2B+%283%2A4%29+%2B+%22...%22+%2B+%282k-1%29+%2A+2k=k%28k%2B1%29%284k-1%29%2F3 ,
lets add %282%28k%2B1%29-1%29+%2A+2%28k%2B1%29 to both sides of the equal sign.
.
Now the left hand side looks like the sum of k%2B1 terms you want.
If you get the right hand side to look like
%28k%2B1%29%28k%2B2%29%284%28k%2B1%29-1%29%2F3=%28k%2B1%29%28k%2B2%29%284k%2B3%29%2F3 ,
you have proven part 2), so it's just a little algebra from now on.
.
.
The sum on the left hand sides has k terms,
If the formula you want to prove is true for n=k ,
those k terms add to k%28k%2B1%29%284k-1%29%2F3 .

%22...%22=%28k%2B1%29%2A%28k%284k-1%29%2B%282k%2B1%29%2A6%29%2F3
%22...%22=%28k%2B1%29%2A%284k%5E2-k%2B12k%2B6%29%2F3
%22...%22=%28k%2B1%29%2A%284k%5E2-k%2B12k%2B6%29%2F3
%22...%22=%28k%2B1%29%2A%284k%5E2%2B11k%2B6%29%2F3
At this point, you realize that %28k%2B2%29%284k%2B3%29=4k%5E2%2B3k%2B8k%2B6=4k%5E2%2B11k%2B6 ,
so it's already really proven that
if the formula you had to prove works for n=k
it must work for n=k%2B1 .
You can continue, writing
%22...%22=%28k%2B1%29%28k%2B2%29%284k%2B3%29%2F3 ,
%22...%22=%28k%2B1%29%28%28k%2B1%29k%2B2%29%284k%2B4-1%29%2F3 and
%22...%22=%28k%2B1%29%28%28k%2B1%29k%2B2%29%284%28k%2B1%29-1%29%2F3 .