Question

The sum of series \[1.2 + 2.3 + 3.4 + ......\]10 terms is
a.440
b.286
c.524
d.\[\infty \]

Answer 1

Hint: This is a general problem of finding the sum of series where we are finding the sum of 10 terms of a given series. We will start with considering the general term, \[{T_{n}} = n\left( {n + 1} \right)\], then proceed generally to find the formula of the sum of the terms in a general form, and then using that we calculate sum of 10 terms.

Complete step-by-step answer:
We have to find,
\[1.2 + 2.3 + 3.4 + ......\]10 terms
The general terms of the above series can be written as,
\[{T_{n}} = n\left( {n + 1} \right)\]
\[\therefore {S_n} = \] ​Sum of n terms of the series.
Now, we get
\[{S_n} = \sum {T_n}\]
\[ = \sum n(n + 1)\]
On Multiplying, we get
\[ = \sum ({n^2} + n)\]
On opening bracket we get,
\[ = \sum {n^2} + \sum n\]
As, sum of squares of 1st n terms, \[ = \dfrac{{n(n + 1)(2n + 1)}}{6}\] and sum of first n terms, \[ = \dfrac{{n(n + 1)}}{2}\], we get
\[ = \dfrac{{n(n + 1)(2n + 1)}}{6} + \dfrac{{n(n + 1)}}{2}\]
On taking \[\dfrac{{n(n + 1)}}{2}\] common we get,
\[ = \dfrac{{n(n + 1)}}{2}[\dfrac{{2n + 1}}{3} + 1]\]
Now, if we simplify, we get
\[ = \dfrac{{n(n + 1)}}{2}[\dfrac{{2n + 1 + 3}}{3}]\]
\[ = \dfrac{{n(n + 1)}}{2}[\dfrac{{2n + 4}}{3}]\]
On taking 2 common from second term,
\[ = \dfrac{{n(n + 1)}}{2}.\dfrac{{2(n + 2)}}{3}\]
On further simplification we get,
\[ = \dfrac{{n(n + 1)(n + 2)}}{3}\]
Hence, \[{S_n} = \dfrac{{n(n + 1)(n + 2)}}{3}\]
As here, we are finding terms of 10 terms, we get, \[n = 10\]
We have, \[{S_n} = \dfrac{{10(10 + 1)(10 + 2)}}{3}\]

On simplification we get,
\[ = \dfrac{{10.11.12}}{3}\]
On division we get,
\[ = 10.11.4\]
On multiplication we get,
\[ = 440\]
Hence, The sum of series \[1.2 + 2.3 + 3.4 + ......\] up to 10 terms is 440
Hence, option (a) is correct.

Note: Here some of the things should be taken care of, that a sum of series of numbers will give us a result only when the series is convergent. If the series is not convergent we will not get a finite sum.
A series is convergent if the sequence of its partial sums tends to a limit; that means that the partial sums become closer and closer to a given number when the number of their terms increases.