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Mathematics Question on Sum of First n Terms of an AP

The value of the sum 1⋅2⋅3+2⋅3⋅4+3⋅4⋅5+…1 \cdot 2 \cdot 3+2 \cdot 3 \cdot 4+3 \cdot 4 \cdot 5+\ldots upto nn terms is equal to

A

16n2(2n2+1)\frac{1}{6} n^{2}\left(2 n^{2}+1\right)

B

16(n2−1)(2n−1)(2n+3)\frac{1}{6}\left(n^{2}-1\right)(2 n-1)(2 n+3)

C

18(n2+1)(n2+5)\frac{1}{8}\left(n^{2}+1\right)\left(n^{2}+5\right)

D

14n(n+1)(n+2)(n+3)\frac{1}{4} n(n+1)(n+2)(n+3)

Answer

14n(n+1)(n+2)(n+3)\frac{1}{4} n(n+1)(n+2)(n+3)

Explanation

Solution

Let given series be
S=1⋅2⋅3+2⋅3⋅4+3⋅4⋅5+…nS=1 \cdot 2 \cdot 3+2 \cdot 3 \cdot 4+3 \cdot 4 \cdot 5+\ldots n term.
Now, nn th terms of the series,
Tn=1+(n−1)⋅12+(n−1)⋅1T_{n}=\\{1+(n-1) \cdot 1\\}\\{2+(n-1) \cdot 1\\}
3+(n−1)⋅1\\{3+(n-1) \cdot 1\\}
[∵Tn=a+(n−1)d]\left[\because T_{n}=a+(n-1) d\right]
=(1+n−1)(2+n−1)(3+n−1)=(1+n-1)(2+n-1)(3+n-1)
=n(n+1)(n+2)=n(n+1)(n+2)
=n(n2+2n+n+2)=n\left(n^{2}+2 n+n+2\right)
=n(n2+3n+2)=n\left(n^{2}+3 n+2\right)
=n3+3n2+2n=n^{3}+3 n^{2}+2 n
Now, S=ΣTn=Σ(n3+3n2+2n)S=\Sigma T_{n}=\Sigma\left(n^{3}+3 n^{2}+2 n\right)
=Σn3+3Σn2+2Σn=\Sigma n^{3}+3 \Sigma n^{2}+2 \Sigma n
=[n(n+1)2]2+3n(n+1)(2n+1)6+2n(n+1)2=\left[\frac{n(n+1)}{2}\right]^{2}+\frac{3 n(n+1)(2 n+1)}{6}+\frac{2 n(n+1)}{2}
=n(n+1)2[n(n+1)2+(2n+1)+2]=\frac{n(n+1)}{2}\left[\frac{n(n+1)}{2}+(2 n+1)+2\right]
=n(n+1)2[n2+n+4n+2+42]=\frac{n(n+1)}{2}\left[\frac{n^{2}+n+4 n+2+4}{2}\right]
=n(n+1)(n2+5n+6)4=\frac{n(n+1)\left(n^{2}+5 n+6\right)}{4}
=n(n+1)(n2+2n+3n+6)4=\frac{n(n+1)\left(n^{2}+2 n+3 n+6\right)}{4}
=n(n+1)[n(n+2)+3(n+2)]4=\frac{n(n+1)[n(n+2)+3(n+2)]}{4}
=n(n+1)(n+2)(n+3)4=\frac{n(n+1)(n+2)(n+3)}{4}