find the number of first n
terms of the series 3 + 7 + 13 + 21 + 31 + so on
Answers
Answer:
Step-by-step explanation:
Let us add 1 in the beginning for the series. We will subtract 1 from the sum of (N+1) terms to get the answer.
Terms are : 1, 3, 7, 13, 21, 31, 43, 57, ..
Differences between terms are : d_n = 2, 4, 6, 8, 12, 14
nth term of the series of differences = d_n = 2 * n.
Sum of this series for n terms = 2 * n(n+1)/2 = n² + n
So the series is :
T1 = 1
T_n+1 = T1 + n² + n , for n >= 1
Rewrite it:
T_n+1 = 1 + n + n² = (n+1)² - (n+1) + 1
=> T_n = n² - n + 1 for n>= 1
So sum of the series for (n+1) terms is now:
= Σ n² - Σ n + Σ 1
= (n+1) (n+2) (2 n + 3) / 6 - (n+1) (n+2)/2 + (n+1)
= (n +1) / 6) [ 2 n² + 7n + 6 - 3 n - 6 + 6 ]
= (n+1) (n² + 2 n + 3) / 3
= (n³ + 3 n² + 5 n + 3) / 3
Sum of the given series for n terms is now : (subtract 1)
= (n³ + 3 n² + 5 n + 3 ) / 3 - 1
= (n³ + 3 n² + 5 n) / 3
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General procedure for such series.
1) Find the series for the differences among terms.
2) Express the n th term in the form of n.
3) Find the sum of n terms of the series of differences.
4) Then add this to the first term T1 of the given series.
5) Now find the sum in terms of Σ n, Σn² etc.