Question 1177107
{{{2}}},{{{5}}},{{{2}}},{{{9}}},{{{2}}},{{{13}}},{{{2}}},{{{17}}},...

pattern:

all odd number terms are equal, {{{2}}}
in {{{100}}} terms you have {{{50}}} of them; so their sum is {{{50*2=100}}}

all even number terms, you also have {{{50}}} of them, have common difference {{{4}}};

{{{5}}}
{{{5+4=9}}}
{{{9+4=13}}}


and so on, and the formula for nth term of an arithmetic progression is 

 {{{a[n]=a[1]+d(n-1)}}} where {{{a[1]=5}}} and {{{d=4}}}
{{{a[n]=5+4(n-1)}}} 
{{{a[n]=5+4n-4}}} 
 so.{{{a[n]=4n+1}}} 

sum of all even number terms is:

{{{S[n]=(n/2)(2a[1]+4(n-1) )}}}
{{{S[50]=(50/2)(2*5+4(50-1) )}}}
{{{S[50]=5150}}}


add both odd and even terms sums 

{{{S[100]=100+5150}}}

{{{S[100]=5250}}}->the sum of the first {{{100}}} terms