Answer :

In the given question we need to find the sum of the series.


For that, first, we need to find the nth term of the series so that we can use summation of the series with standard identities and get the required sum.


The series given is 1 + 5 + 12 + 22 + 35 … to n terms.


This question can be solved by the method of difference.


Note:


Consider a sequence a1, a2, a3 …such that the Sequence a2 –a1, a3 – a2… is either an. A.P. or a G.P.


The nth term, of this sequence, is obtained as follows:


S = a1 + a2 + a3 +…+ an–1 + an (1)


S = a1 + a2 +…+ an–2 + an–1 + an (2)


Subtracting (2) from (1),


We get, an = a1+ [(a2–a1) + (a3–a2) +… (an – an–1)].


Since the terms within the brackets are either in an A.P. or a G.P, we can find the value of an the nth term.


Thus, we can find the sum of the n terms of the sequence as,



So,


By using the method of difference, we can find the nth term of the expression.


Sn = 1 + 5 + 12 + 22 + 35 + ….. + an (1)


Sn = 1 + 5 + 12 + 22+ 35 + …. + an (2)


(1) – (2) 0 = 1 + 4 + 7 + 10+ ….. - an


So, nth term of the series,


an = 1 + 4 + 7 + 10 + ….


So, the nth term form an AP, with the first term, a = 1; common difference, d = 3.


The required nth term of the series is the same as the sum of n terms of AP.


Sum of n terms of an AP,




So, nth term of the series,


Now, we need to find the sum of this series, Sn.




Note:


I. Sum of first n natural numbers, 1 + 2 +3+…n,



II. Sum of squares of first n natural numbers, 12 + 22 + 32+….n2,



III. Sum of cubes of first n natural numbers, 13 + 23 + 33 +…..n3,



IV. Sum of a constant k, N times,



So, for the given series, we need to find,



From, the above identities,






So, Sum of the series,


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