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Question

Find the sum of the following series to n terms:
$$1.2.4+2.3.7+3.4.10+...$$


Solution

Let $$T_n$$ be the terms of the given series. 

Then

$$T_n (n^{th}\ term\ of\ 1,2,3...)\times (n^{th}\ term\ of\ 2,3,4...)\times (n^{th}\ term\ of\ 2,3,4...)$$

$$=[1+(n-1)\times 1].[2+(n-1)\times 1] [4+(n-1)\times 3]$$

$$=[1+n-1][2+n-1][4+3n-3]$$

$$=n (n+1)(3n+1)$$

$$=3n^3+n^2 +3n^2+n$$

$$=3n^3+4n^2 +n$$

$$T_n =3n^3+4n^2+n$$

Let $$S_n$$ denoted sum of $$n$$ terms of the given series 

$$S_n =\displaystyle \sum _{n=1}^{n} T_n =\displaystyle \sum _{n=1}^{n}n^2 +\displaystyle \sum _{n=1}^{n}n$$

$$=\dfrac{3}{1} \left [\dfrac {n(n+1)}{2}\right]^2 +4\left [\dfrac {n(n+1) (2n+1)}{6}\right]+\dfrac {n(n+1)}{2}$$

$$=\dfrac {3}{4}[n(n+1)]^2+\dfrac {2n (n+1) (2n+1)}{3}+\dfrac {n(n+1)}{2}$$

$$=\dfrac {n}{12}(n+1)[9n^2 +9n+16n+8+6]$$

$$S_n =\dfrac {n}{12}(n+1)[9n^2+25n+14]$$

Mathematics

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