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Question

The sum of all possible product of $$1^{st}$$ $$n$$ natural numbers taken two at a time is


A
n(n+1)(n+2)(2n+3)24
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B
n(n21)(3n+2)24
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C
n(n2+1)(3n+2)6
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D
n(n21)(3n+2)6
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Solution

The correct option is A $$\displaystyle \frac{n\left ( n^{2}-1 \right )\left ( 3n+2 \right )}{24}$$
Consider
$$\displaystyle \left ( b_{1}+b_{2}+b_{3}+...+b_{n} \right )^{2}=b_{1}^{2}+b_{2}^{2}+...+b_{n}^{2}+2\sum_{i= j}b_ib_j $$
taking $$\displaystyle b_{1}=1,b_{2}=2,...,b_{n}=n$$
$$\displaystyle \therefore \left ( 1+2+3+...+n \right )^{2}=1^{2}+2^{2}+3^{2}+...+n^{2}+2\displaystyle \sum $$(Product of number taken two at a time)
$$\Rightarrow 2 \sum{b_ib_j} \displaystyle =\left ( 1+2+3+...+n \right )^{2}-\sum_{n=1}^{n}n^{2}\displaystyle=\frac{n^{2}\left ( n+1 \right )^{2}}{4}-\frac{n\left ( n+1 \right
)\left ( 2n+1 \right )}{6}=\frac{n\left ( n^{2}-1 \right )\left ( 3n+2
\right )}{12}$$
$$\Rightarrow \sum{b_ib_j} =\dfrac{n\left ( n^{2}-1 \right )\left ( 3n+2
\right )}{24} $$

Mathematics

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