Question

Given the series:

$2.5+{3.5}^2+{4.5}^3+........+(n+1)5^n$

($T_n$ and $S_n$ represent the $n^{th}$ term and sum of first $n^{t}h$ term)

Find ${}^{\prime }{a}^{\prime }$ and ${}^{\prime }{b}^{\prime }$ such that:

$a{T}_4-b=0$ and $4ab=T_3$

• A. $a=\frac{1}{5},b=5^4$
• B. $a=\frac{1}{5^2},b=5^5$
• C. $a=\frac{4}{5},b=5^6$
• D. None of these

Answer 1

Answer: (A)

$a=\frac{1}{5},b=5^4$

The general term of the series is $(n+1)5^n$

So $T_4=5\times 5^4=3125$ and $T_3=4\times 5^3=500$

Hence the equations become
$3125a-b=0⟹b=3125a$

$4ab=500$

Substituting value of $b,$ we get
$4a\times 3125a=500$

$⟹a^2=\frac{1}{25}$

$⟹a=\frac{1}{5}$

And $b=3125a⟹b=5^4$

Hence, the answer is option A.