I need to know how to proof the arithmetic series below by induction and direct.Please help.

Sum[i=1to n], a_i = n.a_n - sum[i=1 to (n-1)] I{a_(I+1) - a_i}

Hi

To do the induction proof first start by checking the statement is true for n=1.
sum_{i=1}^{1}(a_{i})=a_{1}=1*a_{1}.
Now proceed to the inductive step.
Assume the statement is true for n=k
sum_{i=1}^{k+1}(a_{i})=a_{k+1}+sum_{i=1}^{k}(a_{i} )
=a_{n+1}+ka_{k}-sum_{i=1}^{k-1}[i(a_{i+1}-a_{i})]
from our inductive assumption.
Now add and subtract ka_{k+1} to the left hand side to get
sum_{i=1}^{k+1}(a_{i})=ka_{k+1}-[ka_{k+1}-ka_{k}+sum_{i=1}^{k-1}[i(a_{i+1}-a_{i})]]
Now grouping ka_{k+1}-ka_{k} into the sum we have the result
sum_{i=1}^{k+1}(a_{i})=ka_{k+1}-sum_{i=1}^{k}[i(a_{i+1}-a_{i})]
as required.
To do this by direct proof expand out the left hand side
na_{n}-sum_{i=1}^{n-1}[i(a_{i+1}-a_{i})]=
na_{n}-{(n-1)(a_{n}-a_{n-1})+(n-2)(a_{n-1}-a_{n-2})+... +(n-(n-1))(a_{2}-a_{1})}
=na_{n}-{n(a_{n}-a_{n-1}+a_{n-1}-... -a_{2}+a_{2}-a_{1})-(a_{n}-a_{n-1}+2a_{n-1}-... +(n-1)a_{2}-(n-1)a_{1})
These telescope to give
na_{n}-sum_{i=1}^{n-1}[i(a_{i+1}-a_{i})]=
na_{n}-[na_{n}-na_{1}-sum_{i=1}^{n}(a_{i})+na_{1}]
=sum_{i=1}^{n}(a_{i})
Which is what you need to prove.