Question

find the value of 1×2+2×3+3×4+.......+n(n+1)​

Answer 1

Solution:

We have to derive an expression that gives us the sum of first n terms of this series.

Let us assume the sum is S.

$\displaystyle\rm\longrightarrow S=1\cdot 2+2\cdot 3+3\cdot 4+...+n\cdot(n+1)$

Using sigma notation, it can be written as:

$\displaystyle\rm\longrightarrow S=\sum_{k = 1}^{n}k(k+1)$

On expanding, we get:

$\displaystyle\rm\longrightarrow S=\sum_{k = 1}^{n}(k^2+k)$

$\displaystyle\rm\longrightarrow S=\sum_{k =1}^{n}k^2+\sum_{k = 1}^{n} k$

Now, we know that:

$\bigstar\:\:\underline{\boxed{\rm\sum_{k=1}^{n}k^2=\dfrac{n(n + 1)(2n + 1)}{6}}}$

$\bigstar\:\:\underline{\boxed{\rm\sum_{k=1}^{n}k=\dfrac{n(n + 1)}{2}}}$

Using the above results, we get:

$\displaystyle\rm\longrightarrow S=\dfrac{n(n+1)(2n+1)}{6}+\dfrac{n(n+1)}{2}$

$\displaystyle\rm\longrightarrow S=\dfrac{n(n+1)}{2}\bigg[\dfrac{2n+1}{3}+1\bigg]$

$\displaystyle\rm\longrightarrow S=\dfrac{n(n+1)}{2}\bigg[\dfrac{2n+1+3}{3}\bigg]$

$\displaystyle\rm\longrightarrow S=\dfrac{n(n+1)}{2}\bigg[\dfrac{2n+4}{3}\bigg]$

$\displaystyle\rm\longrightarrow S=\dfrac{n(n+1)(2n+4)}{6}$

$\displaystyle\rm\longrightarrow S=\dfrac{2n(n+1)(n+2)}{6}$

$\displaystyle\rm\longrightarrow S=\dfrac{n(n+1)(n+2)}{3}$

Which is our required answer.

Additional Information:

$\displaystyle1. \: \: \rm \sum^{n}_{i = 1}k = kn$

$\displaystyle2. \: \: \rm \sum^{n}_{i = 1}i = \dfrac{n(n + 1)}{2}$

$\displaystyle3. \: \: \rm \sum^{n}_{i = 1} {i}^{2} = \dfrac{n(n + 1)(2n + 1)}{6}$

$\displaystyle4. \: \: \rm \sum^{n}_{i = 1} {i}^{3} = \bigg[ \dfrac{n(n + 1)}{2} \bigg]^{2}$

$\displaystyle5. \: \: \rm \sum^{n}_{i = 1} 2i = n(n+1)$

$\displaystyle6. \: \: \rm \sum^{n}_{i = 1} \big(2i-1\big)= n^{2}$