find the sum of n terms of the serice s_n =5+11+19+29+41+..........


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Question Number 201253 by mathlove last updated on 02/Dec/23

$f i n d t h e s u m o f n t e r m s o f t h e s e r i c e$ $s_{n} = 5 + 11 + 19 + 29 + 41 + . . . . . . . . . .$
	Answered by Rasheed.Sindhi last updated on 02/Dec/23

	$(4 + 1) + (9 + 2) + (16 + 3) + (25 + 4) + . .$ $(2^{2} + 1) (3^{2} + 2) + (4^{2} + 3) + (5^{2} + 4) + . . .$ $G e n e r a l t e r m : {(k + 1)}^{2} + k$ $= k^{2} + 3 k + 1$ $\underset{k = 1}{\sum^{n}} (k^{2} + 3 k + 1)$ $= \underset{k = 1}{\overset{n}{Σ}} k^{2} + 3 \underset{k = 1}{\overset{n}{Σ}} k + \underset{k = 1}{\overset{n}{Σ}} 1$ $= \frac{n (n + 1) (2 n + 1)}{6} + 3 (\frac{n (n + 1)}{2}) + n$ $= \frac{n (n + 1) (2 n + 1) + 9 n (n + 1) + 6 n}{6}$ $= \frac{n (n + 1) (2 n + 1 + 9) + 6 n}{6}$ $= \frac{n (n + 1) (2 n + 10) + 6 n}{6}$ $= \frac{n (n + 1) (n + 5) + 3 n}{3}$ $= \frac{n (n^{2} + 6 n + 5) + 3 n}{3}$ $= \frac{n (n^{2} + 6 n + 5 + 3)}{3}$ $= \frac{n (n^{2} + 6 n + 8)}{3}$ $= \frac{n (n + 2) (n + 4)}{3}$
	Commented by mathlove last updated on 03/Dec/23

	$t h a n k s$
	Answered by Rasheed.Sindhi last updated on 07/Dec/23
	$s_n =5+11+19+29+41+.......... =(4+1)+(9+2)+(16+3)+(25+4)+... =(4+9+16+...)+(1+2+3+4+...) =(1^2 +2^2 +3^2 +4^2 +..._((n+1) terms) −1)+(1+2+3+4+..._(n terms) ) Formulas: 1+2+3+...+N=((N(N+1))/2) 1^2 +2^2 +3^2 +...+N^2 =((N(N+1)(N+2)(2N+1))/6) =(((n+1)(n+2)(2(n+1)+1))/6)−1+((n(n+1))/2) =(((n+1)(n+2)(2n+3))/6)−1+((n(n+1))/2) =(((n+1)(n+2)(2n+3)−6+3n(n+1))/6) =(((n+1)(n+2)(2n+3)+3n(n+1)−6)/6) =(((n+1){(n+2)(2n+3)+3n}−6)/6) =(((n+1){2n^2 +3n+4n+6+3n}−6)/6) =(((n+1)(2n^2 +10n+6)−6)/6) =((2n^3 +10n^2 +6n+2n^2 +10n+6−6)/6) =((2n^3 +12n^2 +16n)/6) =((n(n^2 +6n+8))/3) =((n(n+2)(n+4))/3) (✓)$
	$s_{n} = 5 + 11 + 19 + 29 + 41 + . . . . . . . . . .$ $= (4 + 1) + (9 + 2) + (16 + 3) + (25 + 4) + . . .$ $= (4 + 9 + 16 + . . .) + (1 + 2 + 3 + 4 + . . .)$ $= (\underset{(n + 1) t e r m s}{1^{2} + 2^{2} + 3^{2} + 4^{2} + . . .} - 1) + (\underset{n t e r m s}{1 + 2 + 3 + 4 + . . .})$ $F o r m u l a s :$ $1 + 2 + 3 + . . . + N = \frac{N (N + 1)}{2}$ $1^{2} + 2^{2} + 3^{2} + . . . + N^{2} = \frac{N (N + 1) (N + 2) (2 N + 1)}{6}$ $= \frac{(n + 1) (n + 2) (2 (n + 1) + 1)}{6} - 1 + \frac{n (n + 1)}{2}$ $= \frac{(n + 1) (n + 2) (2 n + 3)}{6} - 1 + \frac{n (n + 1)}{2}$ $= \frac{(n + 1) (n + 2) (2 n + 3) - 6 + 3 n (n + 1)}{6}$ $= \frac{(n + 1) (n + 2) (2 n + 3) + 3 n (n + 1) - 6}{6}$ $= \frac{(n + 1) {(n + 2) (2 n + 3) + 3 n} - 6}{6}$ $= \frac{(n + 1) {2 n^{2} + 3 n + 4 n + 6 + 3 n} - 6}{6}$ $= \frac{(n + 1) (2 n^{2} + 10 n + 6) - 6}{6}$ $= \frac{2 n^{3} + 10 n^{2} + 6 n + 2 n^{2} + 10 n + 6 - 6}{6}$ $= \frac{2 n^{3} + 12 n^{2} + 16 n}{6}$ $= \frac{n (n^{2} + 6 n + 8)}{3}$ $= \frac{n (n + 2) (n + 4)}{3} (✓)$
	Commented by Rasheed.Sindhi last updated on 07/Dec/23

	$N o w OK!$
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