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Question

Mathematics Question on Differential equations

The sum 1 + 2 ⋅ 3 + 3 ⋅ 32 + …. + 10 ⋅ 39 is equal to

A

2.312+104\frac{2.3^{12}+10}{4}

B

19.310+14\frac{19.3^{10}+1}{4}

C

5.510−25.5^{10}-2

D

9.310+12\frac{9.3^{10}+1}{2}

Answer

19.310+14\frac{19.3^{10}+1}{4}

Explanation

Solution

Let

s=1.3o+2.31+3.32+.......+10.39s=1.3^o+2.3^1+3.3^2+.......+10.3^9

3s=1.31+2.32+.......+10.310{3s=1.3^1+2.3^2+.......+10.3^{10}}

−2s=(1.3o+1.31+1.32+.......+1.39)−10.310-2s=(1.3^o+1.3^1+1.3^2+.......+1.3^9)-10.3^{10}

⇒s=12[10.310−310−13−1]⇒s=\frac{1}{2}[10.3^{10}-\frac{3^{10}-1}{3-1}]

⇒s=12[20.310−310+12]⇒s=\frac{1}{2}[\frac{20.3^{10}{{-3^{10}+1}}}{2}]

⇒S=19.310+14⇒S=\frac{19.3^{10}+1}{4}

So, The correct option is(B): 19.310+14\frac{19.3^{10}+1}{4}