Question

find Sn of the series.
2*5*8+5*8*11+8*11*14+..........​

Answer 1

GIVEN :–

• A Series –

$\\ \implies \bf \: 2.5.8 + 5.8.11 + 8.11.14 +.............$

TO FIND :–

$\\ \implies \bf \: S_n = ?$

SOLUTION :–

$\\ \implies \bf \: 2.5.8 + 5.8.11 + 8.11.14 +.............$

• First sequence in this series –

$\\ \implies \bf \:T_1 = 2 + 5 + 8+.............$

$\\ \implies \bf \:T_1 =2 + (n - 1)3$

$\\ \implies \bf \:T_1 =2 +3n -3$

$\\ \large\implies{ \boxed{ \bf T_1 =3n -1}}$

• Second sequence in this series –

$\\ \implies \bf \:T_2 = 5 + 8+11+.............$

$\\ \implies \bf \:T_2 =5 + (n - 1)3$

$\\ \implies \bf \:T_2 =5+3n -3$

$\\ \large\implies{ \boxed{ \bf T_2 =3n+2}}$

• Third sequence in this series –

$\\ \implies \bf \:T_3 = 8+11+14+.............$

$\\ \implies \bf \:T_3 =8 + (n - 1)3$

$\\ \implies \bf \:T_3 =8+3n -3$

$\\ \large\implies{ \boxed{ \bf T_3 =3n+5}}$

▪︎ So that nth term of given series –

$\\\large\implies{ \boxed{ \bf T_n=(3n -1)(3n + 2)(3n+5)}}$

• We know that –

$\\\implies \bf S_n = \sum T_n$

$\\\implies \bf S_n = \sum (3n -1)(3n + 2)(3n+5)$

$\\\implies \bf S_n = \sum (9 {n}^{2} + 3n - 2)(3n+5)$

$\\\implies \bf S_n = \sum (9 {n}^{2} + 3n - 2)(3n)+(5)(9 {n}^{2} + 3n - 2)$

$\\\implies \bf S_n = \sum 27 {n}^{3} + 9 {n}^{2} - 6n +45{n}^{2} +15n - 10$

$\\\implies \bf S_n = \sum 27 {n}^{3} + 54{n}^{2} + 9n - 10$

$\\\implies \bf S_n = 27\sum{n}^{3} + 54 \sum{n}^{2} + 9 \sum n - 10 \sum (1)$

$\\\implies \bf S_n = 27 \bigg[\dfrac{n(n + 1)}{2} \bigg]^{2} + 54 \bigg[\dfrac{n(n + 1)(2n + 1)}{6} \bigg]+ 9 \bigg[\dfrac{n(n + 1)}{2} \bigg]- 10n$

▪︎Used Identity :–

$\\ \longrightarrow \bf \sum{n}^{3}=\bigg[\dfrac{n(n + 1)}{2} \bigg]^{2}$

$\\ \longrightarrow \bf \sum{n}^{2}=\bigg[\dfrac{n(n + 1)(2n + 1)}{6} \bigg]$

$\\ \longrightarrow \bf \sum n=\bigg[\dfrac{n(n + 1)}{2} \bigg]$

$\\ \longrightarrow \bf \sum(1)=n$

Answer 2

GIVEN :–

• A Series –

$\begin{gathered} \\ \implies \bf \: 2.5.8 + 5.8.11 + 8.11.14 +.............\\ \end{gathered}$

TO FIND :–

$\begin{gathered} \\ \implies \bf \: S_n = ? \\ \end{gathered}$

SOLUTION :–

$\begin{gathered} \\ \implies \bf \: 2.5.8 + 5.8.11 + 8.11.14 +.............\\ \end{gathered}$

• First sequence in this series :–

$\begin{gathered} \\ \implies \bf \:T_1 = 2 + 5 + 8+.............\\ \end{gathered}$

$\begin{gathered} \\ \implies \bf \:T_1 =2 + (n - 1)3 \\ \end{gathered}$

$\begin{gathered} \\ \implies \bf \:T_1 =2 +3n -3 \\ \end{gathered}$

$\begin{gathered} \\ \large\implies{ \boxed{ \bf T_1 =3n -1}}\\ \end{gathered}$

• Second sequence in this series :–

$\begin{gathered} \\ \implies \bf \:T_2 = 5 + 8+11+.............\\ \end{gathered}$

$\begin{gathered} \\ \implies \bf \:T_2 =5 + (n - 1)3 \\ \end{gathered}$

$\begin{gathered} \\ \implies \bf \:T_2 =5+3n -3 \\ \end{gathered}$

$\begin{gathered} \\ \large\implies{ \boxed{ \bf T_2 =3n+2}}\\ \end{gathered}$

• Third sequence in this series :–

$\begin{gathered} \\ \implies \bf \:T_3 = 8+11+14+.............\\ \end{gathered}$

$\begin{gathered} \\ \implies \bf \:T_3 =8 + (n - 1)3 \\ \end{gathered}$

$\begin{gathered} \\ \implies \bf \:T_3 =8+3n -3 \\ \end{gathered}$

$\begin{gathered} \\ \large\implies{ \boxed{ \bf T_3 =3n+5}}\\ \end{gathered}$

▪︎ So that nth term of given series: –

$\begin{gathered} \\\large\implies{ \boxed{ \bf T_n=(3n -1)(3n + 2)(3n+5)}}\\ \end{gathered}$

• We know that :–

$\begin{gathered} \\\implies \bf S_n = \sum T_n\\ \end{gathered}$

$\begin{gathered} \\\implies \bf S_n = \sum (3n -1)(3n + 2)(3n+5)\\ \end{gathered}$

$\begin{gathered} \\\implies \bf S_n = \sum (9 {n}^{2} + 3n - 2)(3n+5)\\ \end{gathered}$

$\begin{gathered} \\\implies \bf S_n = \sum (9 {n}^{2} + 3n - 2)(3n)+(5)(9 {n}^{2} + 3n - 2)\\ \end{gathered}$

$\begin{gathered} \\\implies \bf S_n = \sum 27 {n}^{3} + 9 {n}^{2} - 6n +45{n}^{2} +15n - 10\\ \end{gathered}$

$\begin{gathered} \\\implies \bf S_n = \sum 27 {n}^{3} + 54{n}^{2} + 9n - 10\\ \end{gathered}$

$\begin{gathered} \\\implies \bf S_n = 27\sum{n}^{3} + 54 \sum{n}^{2} + 9 \sum n - 10 \sum (1)\\ \end{gathered}$

$\begin{gathered} \\\implies \bf S_n = 27 \bigg[\dfrac{n(n + 1)}{2} \bigg]^{2} + 54 \bigg[\dfrac{n(n + 1)(2n + 1)}{6} \bigg]+ 9 \bigg[\dfrac{n(n + 1)}{2} \bigg]- 10n\\ \end{gathered}$

▪︎Used Identity :–

$\begin{gathered} \\ \longrightarrow \bf \sum{n}^{3}=\bigg[\dfrac{n(n + 1)}{2} \bigg]^{2}\\ \end{gathered}$

$\begin{gathered} \\ \longrightarrow \bf \sum{n}^{2}=\bigg[\dfrac{n(n + 1)(2n + 1)}{6} \bigg]\\ \end{gathered}$

$\begin{gathered} \\ \longrightarrow \bf \sum n=\bigg[\dfrac{n(n + 1)}{2} \bigg]\\ \end{gathered}$

$\begin{gathered} \\ \longrightarrow \bf \sum(1)=n\\ \end{gathered}$