Question

find the nth term of series
$2 \times 3 + 3 \times 4 + 4 \times 5$
​

Answer 1

$\large\underline{\sf{Solution-}}$

Wᴇ ᴋɴᴏᴡ ᴛʜᴀᴛ,

↝ nᵗʰ term of an arithmetic sequence is,

$\begin{gathered}\red\bigstar\:\:{\underline{\orange{\boxed{\bf{\green{a_n\:=\:a\:+\:(n\:-\:1)\:d}}}}}} \\ \end{gathered}$

Wʜᴇʀᴇ,

• aₙ is the nᵗʰ term.

• a is the first term of the sequence.

• n is the no. of terms.

• d is the common difference.

Now,

Consider,

$\rm :\longmapsto\:2 \times 3 +3 \times 4 + 4 \times 5 + - - - - -$

Thus,

$\bf \: \: \: {n}^{th} \: term \: of \: 2 \times 3 + 3 \times 4 + 4 \times 5 + - - -$

$\sf \: = \: \: ( {n}^{th} \: term \: of \:2, 3, 4, - - - ) \times ( {n}^{th} \: term \: of \: 3, 4, 5, - - - )$

$\sf \: = \: \: \bigg( 2 + (n - 1) \times 1\bigg) \bigg(3 + (n - 1) \times 1 \bigg)$

$\sf \: = \: \: \bigg( 2 + (n - 1)\bigg) \bigg(3 + (n - 1) \bigg)$

$\sf \: = \: \: \bigg( 2 + n - 1\bigg) \bigg(3 + n - 1\bigg)$

$\sf \: = \: \: \bigg( 1 + n \bigg) \bigg(2 + n \bigg)$

$\sf \: = \: \: \bigg(n + 1\bigg) \bigg(n + 2\bigg)$

$\rm\therefore{n}^{th}\:term\:of\: 2 \times 3+3\times 4+4\times5 +- - =(n+1)(n+2)$

Additional Information :-

1. Sum of first n natural numbers is

$\boxed{\bf\: \displaystyle\sum_{r=1}^n \bf\: r \: = \dfrac{n(n + 1)}{2} }$

2. Sum of the squares of first n natural numbers is

$\boxed{\bf\: \displaystyle\sum_{r=1}^n \bf\: {r}^{2} \: = \dfrac{n(n + 1)(2n + 1)}{6}}$

3. Sum of the cubes of first n natural numbers is

$\boxed{\bf\: \displaystyle\sum_{r=1}^n \bf\: {r}^{3} \: = {\bigg(\dfrac{n(n + 1)}{2} \bigg) }^{2} }$