Question

[Math - Sequence and series]

Find the sum of the given series upto n terms.
(1) + (2+3) + (4+5+6) + ...

Here each term in bracket is a separate term.

Answer 1

Hi Friend! here is your answer :

To Find :

the sum of the following sequence till 'n'th number of terms :

1+(2+3)+(4+5+6)+...

Formula To be used :

we will use the formula of total number of terms of a sequence:

if there are x terms in a sequence, then :

total number of terms (whole sum) is given by :

$\displaystyle \rm \frac{x(x+1)}{2}$

Solution :

firstly, we will find out the number of terms in the sequence as from that only we can find the sum of sequence :

$\displaystyle \rm \implies \sum_{k=1}^{n} k = 1+2+3+4+5...+n$

now, using the formula stated above , we get that :

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

now, we know the number of terms in the sequence, so now we will find the sum of the sequence :

$\displaystyle \rm \implies \sum_{k=1}^{\frac{n(n+1)}{2} } k = 1+2+3+4+5+...+\frac{n(n+1)}{2}$

now, as we can clearly see that the sequence so formed is an arithmetic progression, so we will use the formula of sum of AP :

$\displaystyle \rm S = \frac{x}{2} [2a+(x-1)d]$

where, s = sum of AP

x = number of terms of AP

d = common difference of terms

so, applying this formula we get :

${\implies\displaystyle \rm \sum_{k=1}^{\frac{n(n+1)}{2}} k = \frac{n(n+1)}{2\times2} [2(1)+(\frac{n(n+1)}{2} -1)1]}$

${\implies \displaystyle \rm \sum_{k=1}^{\frac{n(n+1)}{2}} k = \frac{n(n+1)}{4} [2-+\frac{n(n+1)}{2} -1]}$

$\implies \displaystyle \rm \sum_{k=1}^{\frac{n(n+1)}{2}} k = \frac{n(n+1)}{4}[\frac{n(n+1)}{2} +1]$

$\implies \displaystyle \rm \sum_{k=1}^{\frac{n(n+1)}{2}} k = \frac{n(n+1)}{4} \times \frac{n^2+n+2}{2}$

$\implies \displaystyle \rm \sum_{k=1}^{\frac{n(n+1)}{2}} k = \frac{n(n+1)(n^2+n+2)}{8}$

$\red{\underline{\boxed{\therefore \displaystyle \rm \sum_{k=1}^{\frac{n(n+1)}{2}} k = \frac{1}{8}n(n+1)(n^2+n+2) }}}$

hence, the sum of the sequence :

$\red{\underline{\boxed{\therefore \displaystyle \rm \sum_{k=1}^{\frac{n(n+1)}{2}} k = \frac{1}{8}n(n+1)(n^2+n+2) }}}$

Answer 2

Please find the above attached image