Question

Find the sum of n terms of the series 1.2 + 2.3 + 3.4 + 4.5 + ....

Answer 1

a = 1.2 
d = 2.3 -1.2 = 1.1 

Sn = n/2(2a + (n-1)d)
    = n/2 (2(1.2) + (n-1)(1.1))
= n/2(2.4 +1.1n - 1.1)
n/2(1.3  + 1.1n)
=(1.3n +1.1n²)2

Answer 2

Answer:

The sum of n terms of series

$\frac{n(n + 1)(n + 2)}{3}$

Step-by-step explanation:

The given series is

$S = 1.2 + 2.3 + 3.4 + .....$

We can write the given series in the following way

$S = 1.(1 + 1) + 2.(2 + 1) + 3(3 + 1) + ....$

From the above expansion,we observe that the nth term of the given series is

$t _{n} = n(n + 1)$

Hence the sum of n terms of series is

$ΣS_{n}=Σn(n+1) \\ = Σ(n {}^{2} + n) \\ = Σ {n}^{2} + Σn$

We know that sum of squares of n natural numbers and sum of n natural numbers.Substituting them in above expression,we get

$ΣS_{n} = \frac{n(n + 1)(2n + 1)}{6} + \frac{n(n + 1)}{2} \\ = \frac{n(n + 1)(n + 2)}{3}$

Therefore,the sum of n terms of series is

$\frac{n(n + 1)(n + 2)}{3}$

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