Question 312694
If you look at the second differences, you'll notice they are the same,
{{{(10-6)=4}}}
{{{(14-10)=4}}}
{{{(18-14)=4}}}
so the generating formula is a quadratic function,
{{{T(n)=an^2+bn+c}}}
Generate the first couple of terms
First term: {{{n=1}}}, {{{a+b+c}}}
Second term:{{{n=2}}}, {{{4a+2b+c}}}
Third term:{{{n=3}}},{{{9a+2b+c}}}
Fourth term:{{{n=4}}},{{{16a+4b+c}}}
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Use the first differences to solve for {{{a}}} and {{{b}}}.
First difference :{{{4a+2b+c-a-b-c=6}}}
1.{{{3a+b=6}}}
Second difference:{{{9a+3b+c-4a-2b-c=10}}}
2.{{{5a+b=10}}}
Subtracting eq. 1 from eq. 2,
{{{5a+b-3a-b=10-6}}}
{{{2a=4}}}
{{{a=2}}}
Then from eq. 1,
{{{3a+b=6}}}
{{{6+b=6}}}
{{{b=0}}}
Then using the first two terms, solve for {{{c}}}.
{{{a+b+c+4a+2b+c=24}}}
{{{2+0+c+8+0+c=24}}}
{{{10+2c=24}}}
{{{2c=14}}}
{{{c=7}}}
So the generating formula is,
{{{T(n)=2n^2+7}}}
{{{highlight( T(1)=9)}}}
{{{highlight( T(2)=15)}}}
{{{highlight( T(3)=25)}}}
{{{T(4)=39}}}
{{{T(5)=57}}}
and you can verify that the first differences are correct.