Find the value of $^{13} C_{1} +^{13} C_{2} + . . . . . . +^{13} C_{13}$ .

Open in App

Solution

We have,
$(1 + x)^{13} =^{13} C_{0} . (1)^{13} . x^{0} +^{13} C_{1} . (1)^{12} . x + . . . . . . . +^{13} C_{13} {.1}^{0} . x^{13}$ .
Putting $x = 1$ both sides we get,
$2^{13} = 1 +$ $^{13} C_{1} +^{13} C_{2} + . . . . . . +^{13} C_{13}$ [Since $^{13} C_{0} = 1$ ]
or, $^{13} C_{1} +^{13} C_{2} + . . . . . . +^{13} C_{13} = 2^{13} - 1$ .

Suggest Corrections

Similar questions

^{13} C_{7} +^{13} C_{8} +^{13} C_{9} +^{13} C_{10} +^{13} C_{11} +^{13} C_{12} +^{13} C_{13}

If(1+x)^n =C0+C1x+C2x^2+......Cnx^n, prove that

1) 3C0-8C1+13C2-18C3+.....upto (n+1) terms

a_{1}, a_{2}, a_{3}, . . . .

a_{1} C_{0} - a_{2} C_{1} + a_{3} C_{2} - . . . . . . + (- 1)^{n} a_{n + 1} C_{n} = 0

3 C_{0} - 8 C_{1} + 13 C_{2} - 18 C_{3} + . . . . = 0

$X = x$	$0$	$1$	$2$
$P (x)$	$4 C^{3}$	$4 C - 13 C^{2}$	$7 C - 1$

C > 0

C =

(666)^{2}

(66666)^{2}

Join BYJU'S Learning Program