Question

Find the sum of the series $1.2+{2.2}^2+{3.2}^3+...+{100.2}^{100}$.

• A. ${100.2}^{101}+2$
• B. ${99.2}^{100}+2$
• C. ${99.2}^{101}+2$
• D. None of these

Answer 1

The correct option is C ${99.2}^{101}+2$

Solve this based on the pattern of the options. The given series has 100 terms. For n = 100. the options can be converted as: Option $\left(a\right)=n\times 2^{(n+1)}+2$. This means that for I term, the sum should be $1\times 2^2+2=6$. But we can see that for I term, the series has a sum of only $1\times 2=2$. Hence, this option can be rejected.
Option (b) = $(n-1)\times 2^n+2$. For 1 term, this gives us a value of 2. For 2 terms, this gives us a value of 6, which does not match the actual value in the question.Hence, this option can be rejected.
Alternatively,
S = $1.2+{2.2}^2+{3.2}^3+...+{100.2}^{100}$ --------- 1)
2*S = ${1.2}^2+{2.2}^3+{3.2}^4+...+{99.2}^{100}+{100.2}^{101}$ ---------- 2)
Subtracting 2) from 1), we get
-S = $2+2^2+2^3+...+2^{100}-{100.2}^{101}$
-S = $2.(2^{100}-1)-{100.2}^{101}$
-S = $2^{101}-2-{100.2}^{101}$
S = ${99.2}^{101}+2$