Question 995635


{{{5}}}, {{{9}}}, {{{13}}},+...+ {{{t[n]}}} has a sum of {{{945}}}



I can see from the sequence {{{5}}}, {{{9}}}, {{{13}}}, ... that each term is {{{4 }}}more than the previous term.

    The first term is {{{5}}}.
    The second term is {{{9=5+4}}}.
    The third term is {{{2}}} fours plus {{{5}}}: {{{13=5+2*4}}}.

Thus the nth term must be {{{n-1}}} fours plus {{{5}}}. 

That is {{{ t[n] = 4(n - 1) + 5 = 1 + 4n}}}.

Thus the {{{n}}} term series is:

    {{{S = 5 + 9 + 13}}} + ... +{{{ (1 + 4(n-1)) + (1 + 4n)}}}

You can find the sum of this series by writing it forward and backwards and then adding down

    {{{S= 5 + 9 }}}+ 	... {{{+ 1+4(n-1) + 1+4n}}}
    {{{S = 1+4n + 1+4(n-1) }}}+ ... + 9 + 5}}}
    {{{2S = 6+4n + 6+4n }}}+ ... + {{{6+4n + 6+4n}}}

Since there are {{{n}}} terms in the series, we have

    {{{2S = n(6 + 4n)}}}

and thus

    {{{S = n(6 + 4n)/2 }}}

since given the sum
{{{ n(6 + 4n)/2 = 945}}} solve for {{{n}}} to find how many terms are in this sequence

{{{ n(6 + 4n) = 945*2}}}

{{{ 6n + 4n^2= 1890}}}

{{{  4n^2+6n -1890=0}}}

{{{n = -45/2}}}....disregard negative solution
{{{n = 21}}}

so, there is {{{21}}} terms in your sequence, and they are

{{{5}}}, {{{9}}}, {{{13}}},{{{17}}}, {{{21}}}, {{{25}}},{{{29}}}, {{{33}}}, {{{37}}},{{{41}}}, {{{45}}}, {{{49}}},{{{53}}}, {{{57}}}, {{{61}}},{{{65}}}, {{{69}}}, {{{73}}},{{{77}}}, {{{81}}}, {{{85}}}

check their sum:

{{{5+9+13+17+21+25+29+33+37+41+45+49+53+57+61+65+69+73+77+81+85=945}}}

{{{945=945}}}