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`1.3+3.5+5.7+......+(2n-1)(2n+1)=(n(4n^2+6n-1))/3` |
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Answer» Let `P9n):1.3+3.5+5.7+....+(2n-1)(2n+1)` `(n(4n^2+6n-1))/(3)` Step I For `n=1`. LHS of Eq. (i) `=1.3=3` RHS of Eq. (i) `=(1(4xx1^2+6xx1-1))/(3)=(4+6-1)/(3)=3` `therefore` LHS = RHS Therefore , P(1) is true. Step II Assume that the result is true for `n=k`. Then, `P(k):1.3+3.5+5.7+.....+(2k-1)(2k+1)=(k(4k^2+6k-1))/(3)` Step III For `n=k+1`, we have to prove that `P(k+1):1.3+3.5+5.7+....+(2k-1)(2k+1)+(2k+1)(2k+3)` `=((k+1)[4(k+1)^2+6(k+1)-1])/(3)` `=((k+1)(4k^2+14k+9))/(3)` LHS `=1.3+3.5+5.7+...+(2k-1)(2k+1)+(2k+1)(2k+3)` `=(k(4k^2+6k-1))/(3)+(2k+1)(2k+3)` [by assumpation step ] `=(4k^3+6k^2-k)/(3)+(4k^2+8k+3)` `=(4k^3+18k^2+23k+9)/(3)` `=((k+1)(4k^2+14k+9))/(3)=RHS` This shows that the result is true for `n=k+1`. Therefore , by the principle of mathematical induction , the result is true for all `n in N`. |
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