Q If ∑r=1n Tr = n8( n + 1) (n + 2) (n + 3), then find â€‹∑r=1n 1Tr - Maths - Sequences and Series

Question

Q If ∑ r = 1 n Tr = n 8 ( n + 1) (n +
2) (n + 3), then find â€‹ ∑ r = 1 n 1 T r

Aarushi Mishra · Accepted Answer

Let Sn=∑r=1nTr=nn+1n+2n+38Sn-1=∑r=1n-1Tr=n-1nn+1n+28____________(1)Sn=∑r=1n-1Tr+Tn=nn+1n+2n+38________________(2)Subtract equation 1 from equation 2Tn=nn+1n+28n+3-n-1=nn+1n+28×4Tn=nn+1n+22Tr=rr+1r+221Tr=2rr+1r+2=r+2-rrr+1r+2=1rr+1-1r+1r+21Tr=1rr+1-1r+1r+21T1=11×2-12×31T2=12×3-13×41T3=13×4-14×5
1Tn=1n×n+1-1n+1×n+2Adding1T1+1T2+1T3+
+1Tn=∑r=1n1Tr=12-1n+1×n+2∑r=1n1Tr=n2+3n+2-22n+1n+2=n2+3n2n+1n+2=nn+32n+1n+2

Q. If ∑ r = 1 n Tr = n 8 ( n + 1) (n + 2) (n + 3), then find ​ ∑ r = 1 n 1 T r .

Q. If $\sum_{r = 1}^{n}$ T_r _{= $\frac{n}{8}$}( n + 1) (n + 2) (n + 3), then find $\sum_{r = 1}^{n}$ $\frac{1}{T_{r}}$ .