Question

The value of $\sum_{1 \leq i < j \leq}^{}{\sum_{n - 1}^{}{(i.j)}}$ⁿC_i .ⁿC_j is equal to –

• A. $\frac{n^{2}}{2}$ (2^2n–2 – ^2n–2C_n–1)
• B. $\frac{n^{2}}{2}$ (2^2n–2 + ^2n–2C_n–1)
• C. n²(2^2n–2 + ^2n–2C_n–1)
• D. n² (2^2n–2 – ^2n–2C_n–1)

Answer 1

Answer: (A)

$\frac{n^{2}}{2}$ (2^2n–2 – ^2n–2C_n–1)

Answer 2

$\sum_{1 \leq i < j \leq}^{}{\sum_{n - 1}^{}{(i.^{n}C_{i})}}$ (j . ⁿC_j)

= $n^{2}\sum_{1 \leq i < j \leq}^{}{\sum_{n - 1}^{}{n - 1C_{i - 1}.^{n - 1}C_{j - 1}}}$

= n².$\left( \frac{2^{2(n - 1)} -^{2(n - 1)}C_{n - 1}}{2} \right)$.