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7.Binomial Theorem
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જો ${C_0},{C_1},{C_2},.......,{C_n}$ એ દ્રીપદી સહગુણક છે , તો $2.{C_1} + {2^3}.{C_3} + {2^5}.{C_5} + ....$ = . . .
A
$\frac{{{3^n} + {{( - 1)}^n}}}{2}$
B
$\frac{{{3^n} - {{( - 1)}^n}}}{2}$
C
$\frac{{{3^n} + 1}}{2}$
D
$\frac{{{3^n} - 1}}{2}$
Solution
(b) ${(1 + x)^n} = {C_0} + {C_1}x + {C_2}{x^2} + {C_3}{x^3} + ….. + {C_n}{x^n}$
${(1 – x)^n} = {C_0} – {C_1}x + {C_2}{x^2} – {C_3}{x^3} + ….. + {( – 1)^n}{C_n}{x^n}$
$[{(1 + x)^n} – {(1 – x)^n}] = 2\,[{C_1}x + {C_3}{x^3} + {C_5}{x^5} + …]$
$\frac{1}{2}[{(1 + x)^n} – {(1 – x)^n}] = {C_1}x + {C_3}{x^3} + {C_5}{x^5} + …….$
Put $x = 2$, $2.{C_1} + {2^3}.{C_3} + {2^5}.{C_5} + …..\, = \frac{{{3^n} – {{( – 1)}^n}}}{2}$
Standard 11
Mathematics
