I know the formula for arithmetic series. What if the difference in the arithmtic series is changing by 1 unit each term.
Can you help me find the general term and sum for:
0+3+7+12+18+25....
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Try looking at the differences of elements $3-0,7-3,12-7,18-12$ can you guess the formula for the sequence of differences? – kingW3 Apr 07 '17 at 15:24
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The difference of the first twp terms is 3. Next terms terms is 4. Next two terms is 5 – Sid Apr 07 '17 at 15:25
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Try summing few terms of the differences of elements what do you get? i.e for $2$ terms you get $3+4$ – kingW3 Apr 07 '17 at 15:30
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This website might be useful for the future. – Surb Apr 07 '17 at 15:34
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http://math.stackexchange.com/questions/88277/how-to-find-nth-term-of-the-sequence-3-7-12-18-25-ldots – Vishnu V.S Apr 07 '17 at 15:45
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Hint: Just as a reminder. This question does not have a unique answer there are infinitely many polynomials satisfying this series.
One easy polynomial is $$a_n=n(n+5)/2=0.5n^2+2.5n$$
MrYouMath
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Just used $n(n+a)/2$ as a test because there seemed to be a quadratic term involved and determined $a$. – MrYouMath Apr 07 '17 at 15:30
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$$ T_2-T_1=3 \\ T_3-T_2=4 \\ ... \\ T_n-T_{n-1}=n+1 $$ Summing them together : $$ T_n-T_1=(n+1)(n+2)-3 \\ T_n=\frac{n^2+3n-4}{2} \\ \sum_{p=1}^n T_i=\sum_{p=1}^n \frac{p^2+3p-4}{2} $$ I'll leave the rest to you.
Vishnu V.S
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