MCQ
The largest perfect square that divides $2014^3-2013^3+2012^3-2011^3+\ldots+2^3-1^3$ is
- A$1^2$
- B$2^2$
- ✓$1007^2$
- D$2014^2$
Let $S=(2014)^3-(2013)^3+(2012)^3$
$-(2011)^3+\ldots+2^3-1^3$
$\Rightarrow S=2\left(2014^3+2012^3+2010^3+\ldots+2^3\right)$
$-\left(2014^3+2013^3+\ldots+2^3+1^3\right)$
$\Rightarrow S=2 \times 2^3\left(1007^3+1006^3+\ldots+1^3\right)$
$-\left(2014^3+2013^3+\ldots+2^3+1^3\right)$
$\Rightarrow S=\frac{2 \times 8(1007)^2(1008)^2 \quad 2014^2 \times 2015^2}{4}$
$\Rightarrow S=(1007)^2\left[\frac{2 \times 8 \times(1008)^2-4 \times(2015)^2}{4}\right]$
$\Rightarrow S=(1007)^2\left[(2016)^2-(2015)^2\right]$
$\Rightarrow S=(1007)^2(2016+2015)(2016-2015)$
$\Rightarrow \quad S=(1007)^2(4031)$
Thus, $S$ is divisible by $(1007)^2$.
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$(x + 2)^{n-1} + (x + 2)^{n-2}. (x + 1) + (x + 2)^{n-3} . (x + 1)^2; + ...... + (x + 1)^{n-1}$ is :
$\operatorname{det}\left[\begin{array}{cc}\sum_{k=0}^n k & \sum_{k=0}^n{ }^n C_k k^2 \\ \sum_{k=0}^n{ }^n C_k k & \sum_{k=0}^n{ }^n C_k 3^k\end{array}\right]=0$, holds for some positive integer $n$. Then $\sum_{k=0}^n \frac{{ }^n C_k}{k+1}$ equals