could someone please help me with the following induction proof. I'm able to complete a somewhat similar proof where the RHS is equal to an algebraic expression ,but, can't get started with the RHS equal to a summation.
$$\sum_{i=1}^n i^3=\left(\sum_{i=1}^n i\right)^2$$
thanks ralph
On
Base case :$n=1$√
Hypothesis: $\sum_{i=1}^{n}i^3=(\sum_{i=1}^{n}i)^2.$
Step: n+1;
$S_n=\sum_{i=1}^{n}i$;
$S_{n+1}=S_n+(n+1)$;
RHS:
$(S_{n+1})^2=(S_n+(n+1))^2=$
$(S_n)^2+2(n+1)S_n+(n+1)^2=$
$(S_n)^2 +(n+1)^2(n)+(n+1)^2=$
$(S_n)^2 +(n+1)^2(n+1)=$
$(S_n)^2+(n+1)^3=$
$\sum_{i=1}^{n}i^3+(n+1)^3= \sum_{i=1}^{n+1}i^3$,
where $(S_n)^2=\sum_{i=2}^{n}i^3$ by hypothesis has been used in the second last expression.
Recall: $S_n= \sum_{i=1}^{n}i =(1/2)(n+1)n$.
Can I presume you know what "proof by induction" is? You "get started" by showing that the statement is true for the 'first case' which, since the sum starts at n= 1, is evaluating both sides for n= 1. The left side is just the single term $1^3= 1$. The right side is single number, 1, squared, which is, of course, 1. They are the same so you have proved the statement for n= 1.
Now, to complete the induction, we need to prove "if the statement is true for n= k then it is also true for n= k+ 1". On the left side, going from n= k to n= k+ 1 means adding $(k+1)^3= k^3+ 3k^2+ 3k+ 1$. Assuming that the statement is true for n= k means assuming that the sum of cubes is the same as $(\sum_{i=1}^k i)^2$ so adding $k^3+ 3k^2+ 3k+ 1$ gives $(\sum_{i=1}^k i)^2+ k^3+ 3k^3+ 3k+ 1$. Is that the same as $(\sum_{i=1}^{k+1} i)^2$?