The numbers $a_n$ are defined by $a_0=1$ and $a_{n+1}=3n^2+n+a_n$ for $n \geq 0$. Then $a_n$ is equal to:

  • A
    $n^3+n^2+1$
  • B
    $n^3-n^2+1$
  • C
    $n^3-n^2$
  • D
    $n^3+n^2$

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What is the sum of $n$ terms of the series $2 + 5 + 14 + 41 + \dots$?

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Let $a_n$ denote the number of all $n$-digit positive integers formed by the digits $0, 1$ or both such that no consecutive digits in them are $0$. Let $b_n$ be the number of such $n$-digit integers ending with digit $1$ and $c_n$ be the number of such $n$-digit integers ending with digit $0$.
$1.$ Which of the following is correct?
$(A)$ $a_{17} = a_{16} + a_{15}$
$(B)$ $c_{17} \neq c_{16} + c_{15}$
$(C)$ $b_{17} \neq b_{16} + c_{16}$
$(D)$ $a_{17} = c_{17} + b_{16}$
$2.$ The value of $b_6$ is
$(A)$ $7$ $(B)$ $8$ $(C)$ $9$ $(D)$ $11$
Give the answer for question $1$ and $2$.

For $x \in \mathbb{R}$,let $[x]$ denote the greatest integer $\le x$. Find the sum of the series $\left[ -\frac{1}{3} \right] + \left[ -\frac{1}{3} - \frac{1}{100} \right] + \left[ -\frac{1}{3} - \frac{2}{100} \right] + \dots + \left[ -\frac{1}{3} - \frac{99}{100} \right]$.

Find the sum of the following series up to $n$ terms:
$\frac{1^{3}}{1}+\frac{1^{3}+2^{3}}{1+3}+\frac{1^{3}+2^{3}+3^{3}}{1+3+5}+\ldots$

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Let $A = \sum_{i=1}^{10} \sum_{j=1}^{10} \min \{i, j\}$ and $B = \sum_{i=1}^{10} \sum_{j=1}^{10} \max \{i, j\}$. Then $A + B$ is equal to

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