Saved Bookmarks
| 1. |
`1^2+2^2+3^2++n^2=(n(n+1)(2n+1))/6` |
|
Answer» Let `P(n):1^2_2^2+3^2+....+n^2=((n+1)(2n+1))/(6)` Step I For n=1 , LHS of Eq. (i) `=1^2=1` RHS of Eq. (i) `=((1+1)(2xx1+1))/(6)` `=(1.2.3)/(6)=1` LHS = RHS Therefore , P(1) is true . Step II Let us assume that the result is true for `n=k`. Then , `P(k):1^2+2^2+3^2+......+k^2=(k(k+1)(2k+1))/(6)` Step III For `n=k+1`, we have to prove that `P(k+1):1^2+2^2+3^2+......+k^2+(k+1)^2` `=((k+1)(k+2)(k+3))/(6)` LHS =`1^2+2^2+3^2+....+k^2+(k+1)^2` `=(k(k+1)(2k+1))/(6)+(k+1)^2` `=(k(k+1)(2k+1))/(6)+(k+1)^2` `=(k+1){(k(2k+1))/(6)+(k+1)}` `=(k+1){(2k^2+7k+6)/(6)}` `=(k+1){((k+2)(2k+3))/(6)}=((k+1)(k+2)(2k+3))/(6)= RHS ` This shows that the result is true for `n=k+1`. Therefore , by the principle of methematical induction ,the result is true for all `n in N`. |
|