MCQ
$\sum\limits_{k = 0}^{10} {^{20}{C_k} = } $
- ✓${2^{19}} + \frac{1}{2}{\,^{20}}{C_{10}}$
- B${2^{19}}$
- C$^{20}{C_{10}}$
- Dએકપણ નહિ.
We know that, ${(1 + x)^n} = {\,^n}{C_0} + {\,^n}{C_1}{x^1} + {\,^n}{C_2}{x^2} + .... + {\,^n}{C_n}.{x^n}$
Put $x = 1$; ${2^n} = {\,^n}{C_0} + {\,^n}{C_1} + {\,^n}{C_2} + ..... + {\,^n}{C_n}$
Put $n = 20$; ${2^{20}} = {\,^{20}}{C_0} + {\,^{20}}{C_1} + {\,^{20}}{C_2} + ...... + {\,^{20}}{C_{20}}$
${2^{20}} + \,{\,^{20}}{C_{10}} = 2\,[{\,^{20}}{C_0} + {\,^{20}}{C_1} + ...... + {\,^{20}}{C_{10}}]$
${[^{20}}{C_0} + {\,^{20}}{C_1} + ...... + {\,^{20}}{C_{10}}] = {2^{19}} + \frac{1}{2}{\,^{20}}{C_{10}}$
$\sum\limits_{k = 0}^{10} {^{20}{C_k}} = {2^{19}} + \frac{1}{2}{\,^{20}}{C_{10}}$.
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| વર્ગ: | $0-6$ | $6-12$ | $12-18$ | $18-24$ | $24-30$ |
| આવૃતિ: | $a$ | $b$ | $12$ | $9$ | $5$ |
જો મધ્યક $=\frac{309}{22}$ અને મધ્યસ્થ $=14$, હોય તો $(a-b)^{2}$ ની કિમંત મેળવો.