Question

Find the sum to n terms of the series 1 × 2 × 3 + 2 × 3 × 4 + 3 × 4 × 5 +.............​

Answer 1

Answer

Given series,

1 × 2 × 3 + 2 × 3 × 4 + 3 × 4 × 5 +.......

nth term t$\sf _n$ = n(n + 1)(n+2)

t$\sf _n$ = n(n² + 3n + 2)

t$\sf _n$ = n³ + n² + 2n

we know,

Sum to n terms = $\sf S_n = \sum\limits_1^nt_n$

$\sf S_n = \sum\limits_1^n( {n}^{3} + 3 {n}^{2} + 2n)$

$\sf S_n = \sum\limits_1^n {n}^{3} + 3\sum\limits_1^n {n}^{2} + 2\sum\limits_1^n n$

$\sf \sum {n}^{3} = \dfrac{ {n}^{2} {(n + 1)}^{2} }{4}$

$\sf \sum {n}^{2} = \dfrac{n(n + 1)(2n + 1)}{6}$

$\sf \sum n = \dfrac{(n + 1)}{2}$

$\sf S_n = \dfrac{ {n}^{2} {(n + 1)}^{2} }{4} + 3 \dfrac{n(n + 1)(2n + 1)}{6} + 2 \dfrac{(n + 1)}{2}$

$\sf S_n = \dfrac{ n {(n + 1)} }{2} \bigg[ \dfrac{n(n + 1)}{2} + 2n + 1 + 2 \bigg]$

$\sf S_n = \dfrac{ n {(n + 1)} }{2} \bigg[ \dfrac{ {n}^{2} + n + 4n + 6 }{2} \bigg]$

$\sf S_n = \dfrac{ n {(n + 1)} }{2} \bigg[ \dfrac{ {n}^{2} +5n + 6 }{2} \bigg]$

$\sf S_n = \dfrac{ n {(n + 1)} }{2} \bigg[ \dfrac{ (n + 2)(n + 3) }{2} \bigg]$

$\sf S_n = \dfrac{ n {(n + 1)(n + 2)(n + 3)} }{4}$