Question 204345
{{{sum(3/(2^i)+4/(5^i),i=1,infinity)}}} Start with the given summation



{{{sum(3(1/(2^i))+4(1/(5^i)),i=1,infinity)}}} Break up the fractions



{{{sum(3((1^i)/(2^i))+4((1^i)/(5^i)),i=1,infinity)}}} Rewrite {{{1}}} as {{{1^i}}}. Note: {{{1^i=1}}} for all values of 'i'



{{{sum(3(1/2)^i+4(1/5)^i,i=1,infinity)}}} Use the identity {{{(x^z)/(y^z)=(x/y)^z}}}



{{{sum(3(1/2)^i,i=1,infinity)+sum(4(1/5)^i,i=1,infinity)}}} Break up the summation.



{{{sum(3(1/2)^(i+1),i=0,infinity)+sum(4(1/5)^(i+1),i=0,infinity)}}} Re-index the summations to start them at zero. Note: this will replace each 'i' with 'i+1' since this offset occurs.



{{{sum(3(1/2)^i,i=0,infinity)-3/(2^0)+sum(4(1/5)^i,i=0,infinity)-4/(5^0)}}} Pull out the first terms of each summation. We're subtracting them off since we're originally starting at i=1 anyway.



{{{sum(3(1/2)^i,i=0,infinity)-3/1+sum(4(1/5)^i,i=0,infinity)-4/1}}} Raise each term to the 0th power to get 1.



{{{sum(3(1/2)^i,i=0,infinity)-3+sum(4(1/5)^i,i=0,infinity)-4}}} Reduce



Now recall that the sum "S" for an infinite geometric series {{{a[n]=a*r^n}}} is {{{S=a/(1-r)}}}. So this means that...



{{{3/(1-1/2)-3+4/(1-1/5)-4}}} Replace the summations with the given sum formulas (see above)



{{{3/(1/2)-3+4/(4/5)-4}}} Subtract



{{{3*(2/1)-3+4*(5/4)-4}}} Multiply by the reciprocal



{{{6/1-3+20/4-4}}} Multiply



{{{6-3+5-4}}} Reduce



{{{4}}} Combine like terms.



So {{{sum(3/(2^i)+4/(5^i),i=1,infinity)=4}}}



Note: if you add up the pieces of this infinite summation, you'll find that the sums will get closer and closer to 4 (you might have to find more terms).