5 ∑k=1n k ( 1+1n) k-1(i) n (n-1) (ii) n (n+1) (iii) n2 (iv) (n+1)2 - Maths - Principle of Mathematical Induction

Question

5   ∑ k = 1 n   k   (   1 + 1 n )  
k - 1 ( i )   n   ( n - 1 )     ( i i )  
n   ( n + 1 )   ( i i i )   n 2     ( i v
)   ( n + 1 ) 2

Lovina Kansal · Accepted Answer

Dear student

We have,Let S=∑k=1nk1+1nk-1↦S=11+1n1-1+21+1n2-1+31+1n3-1+
+n1+1nn-1⇒S=1+21+1n1+31+1n2+ +n1+1nn-1   
(1)Multiply both sides by 1+1n,we get1+1nS=1+1n+21+1n2+31+1n3+
+n1+1nn    
(2)Subtracting (2) from (1), we getS1-1-1n=1+21+1n1+31+1n2+
+n1+1nn-1-1+1n+21+1n2+31+1n3+
+n1+1nn  ⇒-Sn=1+1+1n+1+1n2+
+1+1nn-1-n1+1nn    (3)Consider,1+1+1n+1+1n2+
+1+1nn-1This is a GP with first term,a=1 and common ratio,r=1+1n1=1+1nSo, 1+1+1n+1+1n2+
+1+1nn-1=11+1nn-11+1n-1    ∵Sum of GP=arn-1r-1 if r>1=1+1nn-11nSo, (3) becomes-Sn=1+1nn-11n-n1+1nn⇒-Sn=n1+1nn-n-n1+1nn⇒-Sn--n⇒S=n2
Regards

5 . ∑ k = 1 n k ( 1 + 1 n ) k - 1 ( i ) n ( n - 1 ) ( i i ) n ( n + 1 ) ( i i i ) n 2 ( i v ) ( n + 1 ) 2

${5 . \sum_{k = 1}^{n} k (1 + \frac{1}{n})^{k - 1}}_{}^{} (i) n (n - 1) (i i) n (n + 1) (i i i) n^{2} (i v) (n + 1)^{2}$