∑i=1n∑j=11∑k=1j1 is equal to
n(n+1)2
n(n+1)(n+2)6
n(n+1)22
None of these
Explanation for the correct answer:
Simplifying the given expression:
⇒∑i=1n∑j=11∑k=1j1⇒∑i=1n∑j=11j∵∑k=1j1=1⇒∑i=1n(1+2+3+.....+i)[Considerj=1+2+3+...+i]⇒∑i=1ni(i+1)2∵sumofnnaturalnumbersisn(n+1)2⇒∑i=1ni2+i2⇒n(n+1)(2n+1)12+n(n+1)4∵sumofsquaresofnnaturalnumbers=n(n+1)(2n+1)2⇒n(n+1)42n+13+1⇒n(n+1)(2n+4)4×3⇒n(n+1)(n+2)6
Thus, ∑i=1n∑j=11∑k=1j1=n(n+1)(n+2)6
Hence, option (B) is correct.