Sum to n terms of 1.2.3 + 2.3.4 + 3.4.5 + ……………… is .A) \

1.	Sum to n terms of 1.2.3 + 2.3.4 + 3.4.5 + ……………… is .A) \(\cfrac{n(n+1)(n+2)(n+3)}{6}\) B) \(\cfrac{n(n+1)(n+2)}{2}\)C) \(\cfrac{n(n+1)(n+2)(n+3)}{4}\)D) \(\cfrac{n(n+1)(n+2)(n+3)}{8}\)
Answer» Correct option is (C) \(\frac{n(n+1)(n+2)(n+3)}{4}\) \(S_n=\) 1.2.3 + 2.3.4 + 3.4.5 + …….. + upto n terms = 1.2.3 + 2.3.4 + 3.4.5 + …….. + n (n+1) (n+2) \(=\sum n(n+1)(n+2)\) \(=\sum n(n^2+3n+2)\) \(=\sum n^3+3n^2+2n\) \(=\sum n^3+3\sum n^2+2\sum n\) \(=\left(\frac{n(n+1)}2\right)^2+3\left(\frac{n(n+1)(2n+1)}6\right)+\frac{2n(n+1)}2\) \(=n(n+1)\left(\frac{n(n+1)}4+\frac{2n+1}2+1\right)\) \(=\frac{n(n+1)}4(n^2+n+4n+2+4)\) \(=\frac{n(n+1)}4(n^2+5n+6)\) \(=\frac{n(n+1)(n+2)(n+3)}{4}\) Correct option is C) \(\cfrac{n(n+1)(n+2)(n+3)}{4}\)

Sum to n terms of 1.2.3 + 2.3.4 + 3.4.5 + ……………… is .A) \(\cfrac{n(n+1)(n+2)(n+3)}{6}\) B) \(\cfrac{n(n+1)(n+2)}{2}\)C) \(\cfrac{n(n+1)(n+2)(n+3)}{4}\)D) \(\cfrac{n(n+1)(n+2)(n+3)}{8}\)

Answer»

Correct option is (C) \(\frac{n(n+1)(n+2)(n+3)}{4}\)

\(S_n=\) 1.2.3 + 2.3.4 + 3.4.5 + …….. + upto n terms

= 1.2.3 + 2.3.4 + 3.4.5 + …….. + n (n+1) (n+2)

\(=\sum n(n+1)(n+2)\)

\(=\sum n(n^2+3n+2)\)

\(=\sum n^3+3n^2+2n\)

\(=\sum n^3+3\sum n^2+2\sum n\)

\(=\left(\frac{n(n+1)}2\right)^2+3\left(\frac{n(n+1)(2n+1)}6\right)+\frac{2n(n+1)}2\)

\(=n(n+1)\left(\frac{n(n+1)}4+\frac{2n+1}2+1\right)\)

\(=\frac{n(n+1)}4(n^2+n+4n+2+4)\)

\(=\frac{n(n+1)}4(n^2+5n+6)\)

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

Correct option is C) \(\cfrac{n(n+1)(n+2)(n+3)}{4}\)