# Find the sum of 1st n term of series 3+7+13+21+31--------------

1
by rjsam

2015-10-31T01:11:40+05:30

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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.

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