Q58 / 160 · TS EAMCET 2019 (04 May Shift 2) · Mathematics

For $x=\frac{5}{7}$, if $t_k$ is the first negative term in the expansion of $(1+x)^{7 / 5}$, then, $t_1+t_2+\ldots+t_k=$

  • A
    $\frac{13}{7}$
  • B
    $\frac{107}{14}$
  • Correct answer
    C
    $\frac{104}{49}$
  • D
    $\frac{921}{28}$

Solution

We have, $$ \begin{aligned} (1+x)^{7 / 5}=1 & +\frac{7}{5} x+\frac{1}{2 !} \times \frac{7}{5}\left(\frac{7}{5}-1\right) x^2 \\ & +\frac{1}{3 !}\left(\frac{7}{5}\right)\left(\frac{7}{5}-1\right)\left(\frac{7}{5}-2\right) x^3 \\ t_4 < 0 \quad & t_1+t_2+t_3+t_4 \\ =1 & +\frac{7}{5}\left(\frac{5}{7}\right)+\frac{1}{2} \times \frac{7}{5} \times \frac{2}{5} \times\left(\frac{5}{7}\right)^2 \\ & +\frac{1}{6}\left(\frac{7}{5}\right)\left(\frac{2}{5}\right)\left(\frac{-3}{5}\right)\left(\frac{5}{7}\right)^3\left(\because x=\frac{5}{7}\right) \\ =1 & +1+\frac{1}{7}-\frac{1}{49}=\frac{98+7-1}{49}=\frac{104}{49} \end{aligned} $$

Hence, the correct answer is option: 'C'.

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