Let n be a positive integer. Let A=Σ _k=0^n (-1)^k× Ckn[( 1 2)^k+( 3 4)^k+( 7 8)^k+( 15 16)^k+( 31 32)^k]. If 63A=1- 1 2^30, then n is equal to .

$$A=\sum _{k=0}^{n}Ckn[{(-\frac{1}{2})}^{k}+{(\frac{-3}{4})}^{k}+{(\frac{-7}{8})}^{k}+{(\frac{-15}{16})}^{k}+{(\frac{-31}{32})}^{k}]$$A=k=0∑n​Ckn[(−21​)k+(4−3​)k+(8−7​)k+(16−15​)k+(32−31​)k]

$$A={(1-\frac{1}{2})}^{n}+{(1-\frac{3}{4})}^{n}+{(1-\frac{7}{8})}^{n}+{(1-\frac{15}{16})}^{n}+{(1-\frac{31}{32})}^{n}$$A=(1−21​)n+(1−43​)n+(1−87​)n+(1−1615​)n+(1−3231​)n

$$A=\frac{1}{{2}^{n}}+\frac{1}{{4}^{n}}+\frac{1}{{8}^{n}}+\frac{1}{{16}^{n}}+\frac{1}{{32}^{n}}$$A=2n1​+4n1​+8n1​+16n1​+32n1​

$$A=\frac{1}{{2}^{n}}(\frac{1-{(\frac{1}{{2}^{n}})}^{5}}{1-\frac{1}{{2}^{n}}})\Rightarrow A=\frac{(1-\frac{1}{{2}^{5n}})}{({2}^{n}-1)}$$A=2n1​(1−2n1​1−(2n1​)5​)⇒A=(2n−1)(1−25n1​)​

$$({2}^{n}-1)A=1-\frac{1}{{2}^{5n}},$$(2n−1)A=1−25n1​,

Given, $63A=1-\frac{1}{{2}^{30}}$63A=1−2301​

Clearly $5n=30$5n=30

$$n=6$$n=6