Question

The sum to (n + l) terms of the series c0/2-c1/3+c2/4-c3/5+…is

Answer 1

Do check the attachment you question is solved.

Answer 2

The sum to (n + l) terms of the series c0/2-c1/3+c2/4-c3/5+…

have:S=nC02−nC13+nC24−nC25+...=1

=S=∑nr=0 (−1)rnCrr+2=∑nr=0{ (−1)r (r+1)(r+2)nCr(r+1)}We know that:nCr(r+1)==n+1Cr+1(n+1) ⇒S=∑nr=0{ (−1)r (r+1)(r+2)n+1Cr+1(n+1)}=1(n+1)∑nr=0{ (−1)r (r+1)n+1Cr+1(r+2)}=1(n+1)∑nr=0{ (−1)r (r+1)n+1Cr+1(r+1)+1}

S=1(n+1)∑nr=0{ (−1)r (r+1)n+2Cr+2n+2}=1(n+1)(n+2)∑nr=0{ (−1)r (r+1).n+2Cr+2}=1(n+1)(n+2)∑nr=0{ (−1)r (r+2−1).n+2Cr+2}=1(n+1)(n+2)[∑nr=0{(−1)r (r+2).n+2Cr+2}−∑nr=0{(−1)r .n+2Cr+2}]

We know thatr.nCr=n.n−1Cr−1⇒ (r+2).n+2Cr+2=(n+2).n+1Cr+1⇒S=1(n+1)(n+2)[(n+2)∑nr=0{(−1)r .n+1Cr+1}−∑nr=0{(−1)r .n+2Cr+2}] (i)∑nr=0{(−1)r .n+1Cr+1}=n+1C1−n+1C2+n+1C3−..........={−n+1C0+n+1C1−n+1C2+n+1C3−..........}+n+1C0=(n+1C1+n+1C3+...)−(n+1C0+n+1C2+...)+n+1C0=0+n+1C0 [Sum of odd Binomial Coefficient=Sum of Even Binomial Coefficient]=n+1C0=1∑nr=0{(−1)r .n+2Cr+2}=n+2C2−n+2C3+n+2C4−n+2C5+...={n+2C0−n+2C1+n+2C2−n+2C3+n+2C4−n+2C5+...}+n+2C1−n+2C0=0+n+2C1−n+2C0=n+2−1=n+1Hence (i) becomes:S=1(n+1)(n+2)[(n+2)−(n+1)]=1(n+1)(n+2)[n+2−n−1]=1(n+1)(n+2)⇒nC02−nC13+nC24−nC25+.....=1(n+1)(n+2)