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Question Number 186657 by Mingma last updated on 08/Feb/23

	Answered by Rasheed.Sindhi last updated on 08/Feb/23
	$• Σ_(k=1) ^n T_k : determinant (((T_(n+1) −T_n ),a_n ),((T_2 −T_1 ),a_1 ),((T_3 −T_2 ),a_2 ),((T_4 −T_3 ),a_3 ),((....),(...)),((T_n −T_(n−1) ),a_(n−1) ),((T_(n+1) −T_n ),a_n ),((T_(n+1) −T_1 ),((n/2){2a_1 +(n−1)d}))) T_(n+1) −T_1 =(n/2){2(7)+8(n−1)} T_(n+1) −T_1 =n{7+4(n−1)} T_(n+1) −3=n(4n+3) T_(n+1) =4n^2 +3n+3 T_n =4(n−1)^2 +3(n−1)+3 =4(n^2 −2n+1)+3n−3+3 =4n^2 −8n+4+3n =4n^2 −5n+4 determinant (((T_n =4n^2 −5n+4))) Σ_(k=1) ^n T_k =4Σ_(k=1) ^(n) k^2 −5Σ_(k=1) ^(n) k+4Σ_(k=1) ^(n) 1 =4(((n(n+1)(2n+1))/6))−5(((n(n+1))/2))+4n =((2n(n+1)(2n+1))/3)−((5n(n+1))/2)+4n =((4n(n+1)(2n+1)−15n(n+1)+24n)/6) =n(((8n^2 +12n+4−15n−15+24)/6)) =n(((8n^2 −3n+13)/6)) =((n(8n^2 −3n+13))/6) determinant (((Σ_(k=1) ^n T_k =((n(8n^2 −3n+13))/6)))) • 3T_(n+2) −3T_(n+1) +T_n =3(4(n+2)^2 −5(n+2)+4) −3(4(n+1)^2 −5(n+1)+4) +4n^2 −5n+4 =12(n^2 +4n+4)−15(n+2)+12 −12(n^2 +2n+1)−15(n+1)+12 +4n^2 −5n+4 =12n^2 +48n+48−15n−30+12 −12n^2 −24n−12−15n−15+12 +4n^2 −5n+4 =4n^2 −11n+19$
	$∙ \underset{k = 1}{\sum^{n}} T_{k} :$ $\begin{array}{cccccccc} T_{n + 1} - T_{n} & a_{n} \\ T_{2} - T_{1} & a_{1} \\ T_{3} - T_{2} & a_{2} \\ T_{4} - T_{3} & a_{3} \\ . . . . & . . . \\ T_{n} - T_{n - 1} & a_{n - 1} \\ T_{n + 1} - T_{n} & a_{n} \\ T_{n + 1} - T_{1} & \frac{n}{2} {2 a_{1} + (n - 1) d} \end{array}$ $T_{n + 1} - T_{1} = \frac{n}{2} {2 (7) + 8 (n - 1)}$ $T_{n + 1} - T_{1} = n {7 + 4 (n - 1)}$ $T_{n + 1} - 3 = n (4 n + 3)$ $T_{n + 1} = 4 n^{2} + 3 n + 3$ $T_{n} = 4 {(n - 1)}^{2} + 3 (n - 1) + 3$ $= 4 (n^{2} - 2 n + 1) + 3 n - 3 + 3$ $= 4 n^{2} - 8 n + 4 + 3 n$ $= 4 n^{2} - 5 n + 4$ $\begin{matrix} T_{n} = 4 n^{2} - 5 n + 4 \end{matrix}$ $\underset{k = 1}{\sum^{n}} T_{k} = 4 \underset{k = 1}{\overset{n}{Σ}} k^{2} - 5 \underset{k = 1}{\overset{n}{Σ}} k + 4 \underset{k = 1}{\overset{n}{Σ}} 1$ $= 4 (\frac{n (n + 1) (2 n + 1)}{6}) - 5 (\frac{n (n + 1)}{2}) + 4 n$ $= \frac{2 n (n + 1) (2 n + 1)}{3} - \frac{5 n (n + 1)}{2} + 4 n$ $= \frac{4 n (n + 1) (2 n + 1) - 15 n (n + 1) + 24 n}{6}$ $= n (\frac{8 n^{2} + 12 n + 4 - 15 n - 15 + 24}{6})$ $= n (\frac{8 n^{2} - 3 n + 13}{6})$ $= \frac{n (8 n^{2} - 3 n + 13)}{6}$ $\begin{matrix} \underset{k = 1}{\sum^{n}} T_{k} = \frac{n (8 n^{2} - 3 n + 13)}{6} \end{matrix}$ $∙ 3 T_{n + 2} - 3 T_{n + 1} + T_{n}$ $= 3 (4 {(n + 2)}^{2} - 5 (n + 2) + 4)$ $- 3 (4 {(n + 1)}^{2} - 5 (n + 1) + 4)$ $+ 4 n^{2} - 5 n + 4$ $= 12 (n^{2} + 4 n + 4) - 15 (n + 2) + 12$ $- 12 (n^{2} + 2 n + 1) - 15 (n + 1) + 12$ $+ 4 n^{2} - 5 n + 4$ $= 12 n^{2} + 48 n + 48 - 15 n - 30 + 12$ $- 12 n^{2} - 24 n - 12 - 15 n - 15 + 12$ $+ 4 n^{2} - 5 n + 4$ $= 4 n^{2} - 11 n + 19$
	Commented by Mingma last updated on 12/Feb/23
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