Question

25. Find the sum of the first twelve terms of the series
(3³ - 2³) + (5³-4³) + (7³-6³) + .... to n terms.



$\red{Explanation required}$​

Answer 1

Answer:

$\boxed{\huge{\pink{8280}}}$

Explanation:

$\bold{\underline{\orange{Given }}}\longrightarrow\\ ( {3}^{3} - {2}^{3} ) + ( {5}^{3} - {4}^{3} ) + ( {7}^{3} - {6}^{3} ) + ...........$

$\bold{\underline{\red{To \: find }}}\longrightarrow\\ sum \: of \: first \: 12 \: terms$

$\bold{\underline{\green{Concept \: used}}}/longrightarrow$

$(1)\sum{n} = \dfrac{n(n + 1)}{2} \\ 2) \: \: \sum{ {n}^{2} } = \dfrac{n(n + 1)(2n + 1)}{6} \\ 3) \: \: \sum{ {n}^{3} } = \dfrac{ {n}^{2}(n + 1) ^{2} }{4} \\ 4)if \: we \: take \: first \: member \: of \: each \: term \: \\ 3 \: 5 \: 7....... \\ which \: make \: an \: ap \: whose \\ nth \: term \: = 2n + 1 \\ similarly \: if \: we \: take \: second \: member \\ 2 \: 4 \: 6.... \\ it \: is \: also \: an \: ap \: \\ whose \: nth \: term \: is \: equal \: to \: 2n$

$\bold{\underline{\blue{Solution}}}\longrightarrow \\ let \: nth \: term \: of \: given \: series \: is \: a(n) \\ a(n) = {(2n + 1)}^{3} - {(2n)}^{3} \\ = {(2n)}^{3} + (1) ^{3} + 3 {(2n)}^{2} (1) + 3(1)^{2} (2n) - 8 {n}^{3} \\ = 8 {n}^{3} + 1 + 12 {n}^{2} + 6n - 8 {n}^{3} \\ = 12 {n}^{2} + 6n + 1$

$\sum a(n) = \sum(12 {n}^{2} + 6n + 1)$

$=12 \sum {n}^{2} + 6 \sum \: n + \sum \: 1$

$= 12 \dfrac{n(n + 1)(2n + 1)}{6} + 6 \dfrac{n(n + 1)}{2} + n$

$= 2n(n + 1) \: (2n + 1) + 3n(n + 1) + n$

$= n \: (n + 1)(2(2n + 1) + 3) + n$

$= n(n + 1)(4n + 2 + 3) + n \\ = n(n + 1)(4n + 5) + n$

$now \:for \: sum \: of \: 12 \: terms \\ putting \: n = 12 \: we \: get$

$= 12(12 + 1)(4 \times 12 + 5) + 12 \\ = 12 \times 13(48 + 5) + 12 \\ = 12 \times 13 \times 53 + 12 \\ = 8268 + 12 \\ = 8280$