(Q1) 3 k = an2+ bn. k=1 First term nel , Z k = atb = /. n= 2 2 k = 1+2 =3. KE = a. 2" + 2b = 4a + 2b . atbel . b=-a+ Ly 4a + 2b = 49+ 2(1-a) = 2a + 2 . = 3 . 20 = 3-2 or a = 1 >b=k. Z k = In2 +n ) = In(n+1 ). K=1 Let us prove using induction that this is true I positive integer n , then we have : (1) Case hal , shown as above, also case n= 2. Assume true for n= N, so J K = IN( N+ 1) K=I NOW for heNtl , we have: K + N+1 K= 1 K=I = IN ( N+ 1 ) + (N+ 1) 2 = (N+1) [ N+ 2] 2 true for n= Nt / Ly Hence true for nel, 2, 3