1
JEE Main 2023 (Online) 8th April Morning Shift
+4
-1

Let $$S_{K}=\frac{1+2+\ldots+K}{K}$$ and $$\sum_\limits{j=1}^{n} S_{j}^{2}=\frac{n}{A}\left(B n^{2}+C n+D\right)$$, where $$A, B, C, D \in \mathbb{N}$$ and $$A$$ has least value. Then

A
$$A+B+C+D$$ is divisible by 5
B
$$A+C+D$$ is not divisible by $$B$$
C
$$A+B=5(D-C)$$
D
$$A+B$$ is divisible by $$\mathrm{D}$$
2
JEE Main 2023 (Online) 6th April Evening Shift
+4
-1
Out of Syllabus

If $$\operatorname{gcd}~(\mathrm{m}, \mathrm{n})=1$$ and $$1^{2}-2^{2}+3^{2}-4^{2}+\ldots . .+(2021)^{2}-(2022)^{2}+(2023)^{2}=1012 ~m^{2} n$$ then $$m^{2}-n^{2}$$ is equal to :

A
220
B
200
C
240
D
180
3
JEE Main 2023 (Online) 6th April Morning Shift
+4
-1
Out of Syllabus

The sum of the first $$20$$ terms of the series $$5+11+19+29+41+\ldots$$ is :

A
3420
B
3450
C
3250
D
3520
4
JEE Main 2023 (Online) 1st February Evening Shift
+4
-1
Out of Syllabus

The sum $$\sum\limits_{n = 1}^\infty {{{2{n^2} + 3n + 4} \over {(2n)!}}}$$ is equal to :

A
$${{11e} \over 2} + {7 \over {2e}}$$
B
$${{13e} \over 4} + {5 \over {4e}} - 4$$
C
$${{11e} \over 2} + {7 \over {2e}} - 4$$
D
$${{13e} \over 4} + {5 \over {4e}}$$
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