Question

Compare the sum of n terms of the series: 1 + 2a +3a2 +4a3 +……. and a+ 2a + 3a+4a …up to n terms​

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Answer: $S_{1}$= $\frac{1-a^{n} }{(a-1)^{2} }-\frac{na^{n} }{1-a}$  

              $S_{2}$=$\frac{an(3+n) +2}{2}$

Step-by-step explanation:

• The first series is:-

       S=$1+2a+3a^{2}+4a^{3}+.............(n-1)a^{n-2} +na^{n-1}$

     If we multiply by a on both sides we get,

      Sa=$0+a+2a^{2} +3a^{3}+4a^{4}+.............(n-1)a^{n-1} +na^{n}$

     Subtracting Sa from S we get,

     S(1-a)=$1+a+a^{2}+a^{3}+..............+a^{n-1}-na^{n}$

Now the given series becomes an G.P.

S= $\frac{1-a^{n} }{(a-1)^{2} }-\frac{na^{n} }{1-a}$  is the required sum of the series.

• The second series is:-

        S=1+2a+3a+4a+..........na

We subtract 1 from each sides and get,

       S-1=2a+3a+4a+............na

    ⇒S-1=$\frac{n}{2}[4a+(n-1)a]$

    ⇒ S=$\frac{n}{2}[4a+(n-1)a]$ +1

   ∴ S=$\frac{an(3+n) +2}{2}$

• Hence the equations for sum of the two series are calculated above.

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