1
BITSAT 2024
MCQ (Single Correct Answer)
+3
-1
If $ \sum\limits_{k=1}^{n} k(k+1)(k-1)=p n^{4}+q n^{3}+t n^{2}+s n $, where $ p, q, t $ and $ s $ are constants, then the value of $ s $ is equal to
A
$ -\frac{1}{4} $
B
$ -\frac{1}{2} $
C
$ \frac{1}{2} $
D
$ \frac{1}{4} $
2
BITSAT 2024
MCQ (Single Correct Answer)
+3
-1
There are four numbers of which the first three are in GP and the last three are in AP, whose common difference is 6 . If the first and the last numbers are equal, then two other numbers are
A
$ -2,4 $
B
$ -4,2 $
C
2,6
D
None of the above
3
BITSAT 2024
MCQ (Single Correct Answer)
+3
-1
The coefficient of $ x^{n} $ in the expansion of $ \frac{e^{7 x}+e^{x}}{e^{3 x}} $ is
A
$ \frac{4^{n-1}+(-2)^{n}}{n!} $
B
$ \frac{4^{n-1}+2^{n}}{n!} $
C
$ \frac{4^{n}+(-2)^{n}}{n!} $
D
$ \frac{4^{n-1}+(-2)^{n-1}}{n!} $
4
BITSAT 2023
MCQ (Single Correct Answer)
+3
-1

Let $$\frac{1}{16}, a$$ and $$b$$ be in GP and $$\frac{1}{a}, \frac{1}{b}, 6$$ be in AP, where $$a, b>0$$. Then, $$72(a+b)$$ is equal to

A
12
B
14
C
16
D
18
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