Question

& If the sum to first n terms of a series, the rth term of which is given by (2r + 1)2' can be expressed as
R/
n2) + S2n + T, then find the value of (R + S + T).
T 1
fintacar uuhich are divisible bu

Answer 1

Answer:

The expression can be decomposed as:

(2r+1)2r=r∗2r+1+2r

From those two terms, the second one is a geometric series, easy to sum.

It remains, therefore, the first term:

S1=∑nr=1r∗2r+1=1∗22+2∗23+3∗24+...+n∗2n+1=

=(1+0)∗22+(1+1)∗23+(1+2)∗24+...+(1+[n−1])∗2n+1=

=[22+23+24+...+2n+1]+[1∗23+2∗24+...+(n−1)∗2n+1]

Again, we have two sums, being one of them geometric, and the second has the S1 multiplied by 2. Then, we are close to the end.

S1=2n+2−3−1+2∗S1−2∗n∗2n+1

S1=n∗2n+2−2n+2+4

S=S1+21+22+23+...+2n

S=S1+2n+1−2

S=n∗2n+2−2n+2+4+2n+1−2

S=n∗2n+2−2∗2n+1+2n+1+2

S=n∗2n+2−2n+1+2