If $(3 x - 1) = a_{7} x^{7} + a_{6} x^{6} + . . . . + a_{0}$ , then $a_{7} + a_{6} + . . . + a_{0}$ is

128

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Solution

The correct option is A $128$
$\begin{matrix} {(3 x - 1)}^{7} = [^{7} C_{0} . {(3 x)}^{7} -^{7} C_{1} {(3 x)}^{6} +^{7} C_{2} {(3 x)}^{5} +^{7} C_{3} {(3 x)}^{5} +^{7} C_{4} {(3 x)}^{4} -^{7} C_{5} {(3 x)}^{5} +^{7} C_{6} {(3 x)}^{6} -^{7} C_{7} {(3 x)}^{7}] \Rightarrow 2187 x^{7} - 5103 x^{6} + 5103 x^{5} - 2035 x^{4} + 945 x^{3} - 189 x^{2} + 21 x - 1 [\begin{matrix} ^{n} C_{1} = n^{n} C_{0} =^{n} C_{n} = 1 \end{matrix}] \Rightarrow∴ a_{7} + a_{6} + a_{5} + a_{4} + . . . . . . . . . . . . a_{0} = 2187 - 5103 + 5103 - 2835 + 945 - 189 + 21 - 1 = 128 \end{matrix}$

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Similar questions

Q. If

(3 x - 1)^{7} = a_{7} x^{7} + a_{6} x^{6} + a_{5} x^{5} + . . . + a_{0}

(A) then

a_{7} + a_{6} + . . . a_{0} = ?

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(d) 64

The equation $a_{8} x^{8} + a_{7} x^{7} + a_{6} x^{6} + . . . + a_{0} = 0$ has all its roots positive and real $(w h e r e a_{8} = 1, a_{7} = - 4, a_{0} = {\frac{1}{2}}^{8})$ , then

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a_{8} X^{8} + a_{7} X^{7} + a_{6} X^{6} + \dots + a_{0} = 0

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(w h e r e a_{8} = 1, a_{7} = - 4, a_{0} = \frac{1}{2^{8}})

, then which of the following is/are true?

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If (3x−1)=a7x7+a6x6+....+a0, then a7+a6+...+a0 is

The correct option is A 128(3x−1)7=[7C0.(3x)7−7C1(3x)6+7C2(3x)5+7C3(3x)5+7C4(3x)4−7C5(3x)5+7C6(3x)6−7C7(3x)7]⇒2187x7−5103x6+5103x5−2035x4+945x3−189x2+21x−1[nC1=nnC0=nCn=1]⇒∴a7+a6+a5+a4+............a0=2187−5103+5103−2835+945−189+21−1=128

If $(3 x - 1) = a_{7} x^{7} + a_{6} x^{6} + . . . . + a_{0}$ , then $a_{7} + a_{6} + . . . + a_{0}$ is