a-n-in-numerical-series-7-9-11-13-2n-1-an-2-bn-c-find-a-b-c-

Question Number 157908 by HongKing last updated on 29/Oct/21

(a_{n}) in numerical series

7 + 9 + 11 + 13 + \dots + (2 n + 1) = {an}^{2} + bn + c

find a + b + c = ?

Answered by Rasheed.Sindhi last updated on 29/Oct/21

n starts from 3

for n = 3 :

7 = a {(3)}^{2} + b (3) + c

9 a + 3 b + c = 7 \dots \dots \dots \dots \dots \dots (i)

for n = 4 :

7 + 9 = a {(4)}^{2} + b (4) + c

16 a + 4 b + c = 16 \dots \dots \dots \dots \dots . . (ii)

(ii) - (i) : 7 a + b = 9 \dots \dots \dots \dots . . (iii)

for n = 5

7 + 9 + 11 = a {(5)}^{2} + 5 b + c

25 a + 5 b + c = 27 \dots \dots \dots \dots \dots . . (iv)

(iv) - (ii) : 9 a + b = 11 \dots \dots \dots . (v)

(v) - (iii) : 2 a = 2 \Rightarrow a = 1

(iii) \Rightarrow b = 9 - 7 a = 9 - 7 (1) = 2

(i) : 9 a + 3 b + c = 7 \Rightarrow 9 (1) + 3 (2) + c = 7

\Rightarrow c = 7 - 15 = - 8

a = 1, b = 2, c = - 8

a + b + c = - 5

Commented by HongKing last updated on 29/Oct/21

alot thank you so much sir