Question

Find the sum of the series \[\sum\limits_{r = 0}^n {{{\left( { - 1} \right)}^r}{.^n}{C_r}\left[ {\dfrac{1}{{{2^r}}} + \dfrac{{{3^r}}}{{{2^{2r}}}} + \dfrac{{{7^r}}}{{{2^{3r}}}} + \dfrac{{{{15}^r}}}{{{2^{4r}}}} + } \right.} ...\] up to m terms

Answer 1

Hint: Since we have to find the series of the above summation, try to first open the given bracket. After this you will get a series of terms where you have to apply related binomial formulas to individual terms. After further solving this the series generated will be geometric and hence it’s sum can be found out using the geometric series formula.

Complete step-by-step answer:
we will first try to note down what is given to us.
We have to find the sum of this series \[\sum\limits_{r = 0}^n {{{\left( { - 1} \right)}^r}{.^n}{C_r}\left[ {\dfrac{1}{{{2^r}}} + \dfrac{{{3^r}}}{{{2^{2r}}}} + \dfrac{{{7^r}}}{{{2^{3r}}}} + \dfrac{{{{15}^r}}}{{{2^{4r}}}} + } \right.} ...\] m terms.
We will start the solution by opening the bracket. Therefore we get
 \[\sum\limits_{r = 0}^n {{{\left( { - 1} \right)}^r}{.^n}{C_r}\left[ {\dfrac{1}{{{2^r}}} + \dfrac{{{3^r}}}{{{2^{2r}}}} + \dfrac{{{7^r}}}{{{2^{3r}}}} + \dfrac{{{{15}^r}}}{{{2^{4r}}}} + } \right.} ...\] m terms.
= \[\sum\limits_{r = 0}^n {{{\left( { - 1} \right)}^r}{.^n}{C_r}\dfrac{1}{{{2^r}}} + \sum\limits_{r = 0}^n {{{\left( { - 1} \right)}^r}{.^n}{C_r}} \dfrac{{{3^r}}}{{{2^{2r}}}} + \sum\limits_{r = 0}^n {{{\left( { - 1} \right)}^r}{.^n}{C_r}} \dfrac{{{7^r}}}{{{2^{3r}}}} + } ...\] up to m terms
Solving further, we get:
= \[\sum\limits_{r = 0}^n {{{\left( { - 1} \right)}^r}{.^n}{C_r}{{\left( {\dfrac{1}{2}} \right)}^r} + \sum\limits_{r = 0}^n {{{\left( { - 1} \right)}^r}{.^n}{C_r}} {{\left( {\dfrac{3}{4}} \right)}^r} + \sum\limits_{r = 0}^n {{{\left( { - 1} \right)}^r}{.^n}{C_r}} {{\left( {\dfrac{7}{8}} \right)}^r} + } ...\] up to m terms
Now since we know that \[\sum\limits_{r = 0}^n {{{\left( { - 1} \right)}^r}{.^n}{C_r}.{x^r} = {{\left( {1 - x} \right)}^n}} \] , using this formula in the above equation, we get
 \[\sum\limits_{r = 0}^n {{{\left( { - 1} \right)}^r}{.^n}{C_r}{{\left( {\dfrac{1}{2}} \right)}^r} + \sum\limits_{r = 0}^n {{{\left( { - 1} \right)}^r}{.^n}{C_r}} {{\left( {\dfrac{3}{4}} \right)}^r} + \sum\limits_{r = 0}^n {{{\left( { - 1} \right)}^r}{.^n}{C_r}} {{\left( {\dfrac{7}{8}} \right)}^r} + } ...\] up to m terms
= \[{\left( {1 - \dfrac{1}{2}} \right)^n} + {\left( {1 - \dfrac{3}{4}} \right)^n} + {\left( {1 - \dfrac{7}{8}} \right)^n} + ...\] up to m terms
Solving further we get
 \[{\left( {1 - \dfrac{1}{2}} \right)^n} + {\left( {1 - \dfrac{3}{4}} \right)^n} + {\left( {1 - \dfrac{7}{8}} \right)^n} + ...\] up to m terms
= \[{\left( {\dfrac{1}{2}} \right)^n} + {\left( {\dfrac{1}{4}} \right)^n} + {\left( {\dfrac{1}{8}} \right)^n} + ...\] up to m terms
Now the above obtained series is a geometric series with terms up to m.
hence we have to use the formula to find the sum of geometric series which is ${S_n} = \dfrac{{a\left( {1 - {r^n}} \right)}}{{\left( {1 - r} \right)}}$
where a is the first term and r is the common ratio.
Hence comparing to our given series we get $a = {\left( {\dfrac{1}{2}} \right)^r}$ and $r = \dfrac{1}{{{2^n}}}$
Therefore we get
 \[{\left( {\dfrac{1}{2}} \right)^n} + {\left( {\dfrac{1}{4}} \right)^n} + {\left( {\dfrac{1}{8}} \right)^n} + ...\] up to m terms = $\dfrac{{{{\left( {\dfrac{1}{2}} \right)}^n}\left( {1 - {{\left( {\dfrac{1}{{{2^n}}}} \right)}^m}} \right)}}{{\left( {1 - \dfrac{1}{{{2^n}}}} \right)}}$
=$\dfrac{{{2^{mm}} - 1}}{{{2^{mm}}\left( {{2^n} - 1} \right)}}$
Hence sum of the series \[\sum\limits_{r = 0}^n {{{\left( { - 1} \right)}^r}{.^n}{C_r}\left[ {\dfrac{1}{{{2^r}}} + \dfrac{{{3^r}}}{{{2^{2r}}}} + \dfrac{{{7^r}}}{{{2^{3r}}}} + \dfrac{{{{15}^r}}}{{{2^{4r}}}} + } \right.} ...\] up to m terms is $\dfrac{{{2^{nm}} - 1}}{{{2^{nm}}\left( {{2^n} - 1} \right)}}$ respectively.
So, the correct answer is “$\dfrac{{{2^{nm}} - 1}}{{{2^{nm}}\left( {{2^n} - 1} \right)}}$”.

Note: In this kind of problems try to first simplify the question and get it in the form of a simplified series. This will help you to apply any formulas if available to the individual terms of the series will make the series more simplified. The binomial theorem helps us in these cases to make the large polynomial compact.