SATHEE: Principle Of Mathematical Induction Question 16

Principle Of Mathematical Induction Question 16

Question: Sum of $ \frac{1^{3}}{1}+\frac{1^{3}+2^{3}}{1+2}+… $ to n terms is

Options:

A) $ \frac{(n+1)(n+2)}{3} $

B) $ n(n+1)(n+2) $

C) $ \frac{n(n+1)(n+2)}{6} $

D) $ \frac{n(n+1)(n+2)}{2} $

Show Answer

Answer:

Correct Answer: C

Solution:

[c] $ S=\sum{T_{n}=\sum{\frac{1^{3}+2^{3}+…+n^{3}}{1+2+…+n}}=\sum{\frac{\sum{n^{3}}}{\sum{n}}}} $ $ =\sum{\frac{\frac{n^{2}{{(n+1)}^{2}}}{4}}{\frac{n(n+1)}{2}}}=\sum{\frac{n(n+1)}{2}=\frac{1}{2}[\sum{n^{2}+\sum{n}}]} $ $ =\frac{1}{2}[ \frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2} ] $ $ =\frac{n(n+1)}{4}[ \frac{2n+1}{3}+1 ]=\frac{n(n+1)(2n+4)}{12} $ $ =\frac{n(n+1)(n+2)}{6}. $

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Principle Of Mathematical Induction Question 16

Question: Sum of $ \frac{1^{3}}{1}+\frac{1^{3}+2^{3}}{1+2}+… $ to n terms is

Options:

Answer:

Solution:

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