- A215 - 15
- B216 - 16
- C216 - 17
- D217 - 17
Solution:
Consider given the binomial expression,
$\sum\limits^{16}_{\text{r}+2}{^{16}}\text{C}_\text{r}={^{16}}\text{C}_{2}+{^{16}}\text{C}_{3}+{^{16}}\text{C}_{3}+\ .....\ {^{16}}\text{C}_{16}$
$=2^{16}-17$
Hence, this is the answer.
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If rth term is the middle term in the expansion of $\Big(\text{x}^{2}-\frac{1}{2\text{x}}\Big)^{20},$ then (r + 3)th term is:
${^\text{20}}\text{C}_{\text{14}}\ \Big(\frac{\text{x}}{2^{14}}\Big)$
${^\text{20}}\text{C}_{\text{12}}\ \text{x}^{2}\ 2^{-12}$
$-{^\text{20}}\text{C}_{\text{7}}\ \text{x}\ 2^{-13}$
None of these.
If $\cos\text{x}=\frac{1}{2}\Big(\text{a}+\frac{1}{\text{a}}\Big),$ and $3\text{x}=\lambda\Big(\text{a}^3+\frac{1}{\text{a}^3}\Big),$ then $\lambda=$
$\frac{1}{4}$
$\frac{1}{2}$
$1$